Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89<\/strong>: <\/p>\n\n\n\na) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3.<\/ins><\/p>\n\n\n\nb) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb<\/p>\n\n\n\nTr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x;<\/p>\n\n\n\nC\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x.<\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\na)-2x + 3 > 0 \u21d4 -2x > -3 \u21d4 x < 3\/2<\/p>\n\n\n\nBi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m tr\u00ean tr\u1ee5c s\u1ed1:<\/p>\n\n\n\n<\/figure>\n\n\n\nb) Nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb:<\/p>\n\n\n\nTr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x < 3\/2<\/p>\n\n\n\nC\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x > 3\/2<\/p>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 90<\/strong>: X\u00e9t d\u1ea5u c\u00e1c nh\u1ecb th\u1ee9c f(x) = 3x + 2, g(x) = -2x + 5.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: X\u00e9t d\u1ea5u bi\u1ec3u th\u1ee9c f(x) = (2x \u2013 1)(-x + 3)<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3.<\/ins><\/p>\n\n\n\nb) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb<\/p>\n\n\n\nTr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x;<\/p>\n\n\n\nC\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x.<\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\na)-2x + 3 > 0 \u21d4 -2x > -3 \u21d4 x < 3\/2<\/p>\n\n\n\nBi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m tr\u00ean tr\u1ee5c s\u1ed1:<\/p>\n\n\n\n<\/figure>\n\n\n\nb) Nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb:<\/p>\n\n\n\nTr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x < 3\/2<\/p>\n\n\n\nC\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x > 3\/2<\/p>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 90<\/strong>: X\u00e9t d\u1ea5u c\u00e1c nh\u1ecb th\u1ee9c f(x) = 3x + 2, g(x) = -2x + 5.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: X\u00e9t d\u1ea5u bi\u1ec3u th\u1ee9c f(x) = (2x \u2013 1)(-x + 3)<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb<\/p>\n\n\n\n
Tr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x;<\/p>\n\n\n\n
C\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x.<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\na)-2x + 3 > 0 \u21d4 -2x > -3 \u21d4 x < 3\/2<\/p>\n\n\n\nBi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m tr\u00ean tr\u1ee5c s\u1ed1:<\/p>\n\n\n\n<\/figure>\n\n\n\nb) Nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb:<\/p>\n\n\n\nTr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x < 3\/2<\/p>\n\n\n\nC\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x > 3\/2<\/p>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 90<\/strong>: X\u00e9t d\u1ea5u c\u00e1c nh\u1ecb th\u1ee9c f(x) = 3x + 2, g(x) = -2x + 5.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: X\u00e9t d\u1ea5u bi\u1ec3u th\u1ee9c f(x) = (2x \u2013 1)(-x + 3)<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
a)-2x + 3 > 0 \u21d4 -2x > -3 \u21d4 x < 3\/2<\/p>\n\n\n\n
Bi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m tr\u00ean tr\u1ee5c s\u1ed1:<\/p>\n\n\n\n<\/figure>\n\n\n\nb) Nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb:<\/p>\n\n\n\nTr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x < 3\/2<\/p>\n\n\n\nC\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x > 3\/2<\/p>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 90<\/strong>: X\u00e9t d\u1ea5u c\u00e1c nh\u1ecb th\u1ee9c f(x) = 3x + 2, g(x) = -2x + 5.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: X\u00e9t d\u1ea5u bi\u1ec3u th\u1ee9c f(x) = (2x \u2013 1)(-x + 3)<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
b) Nh\u1ecb th\u1ee9c f(x) = -2x + 3 c\u00f3 gi\u00e1 tr\u1ecb:<\/p>\n\n\n\n
Tr\u00e1i d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x < 3\/2<\/p>\n\n\n\n
C\u00f9ng d\u1ea5u v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a x khi x > 3\/2<\/p>\n\n\n\n
Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 90<\/strong>: X\u00e9t d\u1ea5u c\u00e1c nh\u1ecb th\u1ee9c f(x) = 3x + 2, g(x) = -2x + 5.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: X\u00e9t d\u1ea5u bi\u1ec3u th\u1ee9c f(x) = (2x \u2013 1)(-x + 3)<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: X\u00e9t d\u1ea5u bi\u1ec3u th\u1ee9c f(x) = (2x \u2013 1)(-x + 3)<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: X\u00e9t d\u1ea5u bi\u1ec3u th\u1ee9c f(x) = (2x \u2013 1)(-x + 3)<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nTr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 92<\/strong>: Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup> \u2013 4x < 0.<\/ins><\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\nx3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
x3<\/sup> \u2013 4x < 0 \u21d4 x(x2<\/sup> – 4) < 0 \u21d4 x(x – 2)(x + 2) < 0<\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n\n\n\n<\/figure>\n\n\n\nT\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\nS = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\nB\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
T\u1eeb b\u1ea3ng x\u00e9t d\u1ea5u ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/p>\n\n\n\n
S = (-\u221e;2) \u222a (0;2)<\/p>\n\n\n\n
B\u00e0i 1 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: X\u00e9t d\u1ea5u c\u00e1c bi\u1ec3u th\u1ee9c:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
a) Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u00e0 1\/2 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\n
Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\nb) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
K\u1ebft lu\u1eadn : <\/p>\n\n\n\n
+ f(x) > 0 khi x < \u20133 ho\u1eb7c x > 1\/2<\/p>\n\n\n\n
+ f(x) < 0 khi \u20133 < x < 1\/2<\/p>\n\n\n\n
+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = 1\/2. <\/p>\n\n\n\n
b) Nh\u1ecb th\u1ee9c \u20133x \u2013 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20131; nh\u1ecb th\u1ee9c x + 2 c\u00f3 nghi\u1ec7m l\u00e0 \u20132 ; nh\u1ecb th\u1ee9c x + 3 c\u00f3 nghi\u1ec7m l\u00e0 \u20133. <\/p>\n\n\n\n
Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u : <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
+ f(x) < 0 khi \u20133 < x < \u20132 ho\u1eb7c x > \u20131<\/p>\n\n\n\n
+ f(x) > 0 khi x < \u20133 ho\u1eb7c \u20132 < x < \u20131. <\/p>\n\n\n\n
+ f(x) = 0 khi x = \u20133 ho\u1eb7c x = \u20132 ho\u1eb7c x = \u20131. <\/ins><\/p>\n\n\n\nc) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
c) Ta c\u00f3: <\/p>\n\n\n\n<\/figure>\n\n\n\nNh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
Nh\u1ecb th\u1ee9c \u20135x \u2013 11 c\u00f3 nghi\u1ec7m l\u00e0 \u201311\/5, nh\u1ecb th\u1ee9c 3x +1 c\u00f3 nghi\u1ec7m l\u00e0 \u20131\/3, nh\u1ecb th\u1ee9c 2 \u2013 x c\u00f3 nghi\u1ec7m l\u00e0 2. <\/p>\n\n\n\n
Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\nd) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
+ f(x) > 0 khi \u201311\/5 < x < \u20131\/3 ho\u1eb7c x > 2. <\/p>\n\n\n\n
+ f(x) < 0 khi x < \u201311\/5 ho\u1eb7c \u20131\/3 < x < 2. <\/p>\n\n\n\n
+ f(x) = 0 khi x = \u201311\/5. <\/p>\n\n\n\n
+ Khi x = \u20131\/3 ho\u1eb7c x = 2, f(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\n
d) f(x) = 4x2<\/sup> \u2013 1 = (2x \u2013 1)(2x + 1) <\/p>\n\n\n\nNh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
Nh\u1ecb th\u1ee9c 2x \u2013 1 c\u00f3 nghi\u1ec7m x = 1\/2, nh\u1ecb th\u1ee9c 2x + 1 c\u00f3 nghi\u1ec7m x = \u20131\/2. <\/p>\n\n\n\n
Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u: <\/p>\n\n\n\n<\/figure>\n\n\n\nK\u1ebft lu\u1eadn : <\/p>\n\n\n\n+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\nB\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
+ f(x) > 0 khi x < \u20131\/2 ho\u1eb7c x > 1\/2. <\/p>\n\n\n\n
+ f(x) < 0 khi \u20131\/2 < x < 1\/2<\/p>\n\n\n\n
+ f(x) = 0 khi x = 1\/2 ho\u1eb7c x = \u20131\/2.<\/p>\n\n\n\n
B\u00e0i 2 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n<\/figure>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\na) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
a) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nC\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
C\u00e1c nh\u1ecb th\u1ee9c \u2013x + 3; x \u2013 1; 2x \u2013 1 c\u00f3 nghi\u1ec7m l\u1ea7n l\u01b0\u1ee3t l\u00e0 3; 1; 1\/2.<\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
D\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u th\u1ea5y <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\n
b) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1 v\u00e0 x \u2260 \u20131. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
\n (Nh\u00e2n c\u1ea3 hai v\u1ebf c\u1ee7a B\u0110T v\u1edbi (x \u2013 1)2 > 0). <\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
D\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
V\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m S = (\u2013\u221e; \u20131] \u222a (0; 3)\\{1}<\/ins><\/p>\n\n\n\nc) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
c) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 0; x \u2260 \u20133; x \u2260 \u20134. <\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
\u0110\u1eb7t \n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
D\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\nd) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S = (\u201312; \u20134) \u222a (\u20133; 0). <\/p>\n\n\n\n
d) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 \u00b11.<\/p>\n\n\n\n<\/figure>\n\n\n\n\u0110\u1eb7t<\/figcaption><\/figure>\n\n\n\n. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
. Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
D\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y \n<\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
V\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 <\/p>\n\n\n\n<\/figure>\n\n\n\nB\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
B\u00e0i 3 (trang 94 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\na) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
a) |5x – 4| \u2265 6<\/ins><\/p>\n\n\n\n<\/figure>\n\n\n\nb) <\/p>\n\n\n\nL\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
b) <\/p>\n\n\n\n
L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n<\/figure>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
V\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
<\/ins><\/p>\n\n\n\nb) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
b) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2260 1; x \u2260 \u20132. <\/p>\n\n\n\n<\/figure>\n\n\n\nTa c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u sau: <\/p>\n\n\n\n<\/figure>\n\n\n\nD\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\nV\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\nKi\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
D\u1ef1a v\u00e0o b\u1ea3ng x\u00e9t d\u1ea5u ta th\u1ea5y (x + 1)(x + 5) > 0 khi \u20135 < x < \u20131. <\/p>\n\n\n\n
V\u1eady b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m (\u20135; \u20131)\\{\u20132}. <\/p>\n\n\n\n
Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n |f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n |f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
+ V\u1edbi a > 0 ta c\u00f3: <\/p>\n\n\n\n
|f(x)| \u2264 a \u21d4 \u2013a \u2264 f(x) \u2264 a. <\/p>\n\n\n\n
|f(x)| \u2265 a \u21d4 f(x) \u2265 a ho\u1eb7c f(x) \u2264 \u2013a <\/p>\n\n\n\n
+ f2<\/sup>(x) \u2264 g2<\/sup>(x) \u21d4 |f(x)| \u2264 |g(x)| v\u00ec \n<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 3: D\u1ea5u c\u1ee7a nh\u1ecb th\u1ee9c b\u1eadc nh\u1ea5t<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n\n\n\n\t\n\t\n\t\n
Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 89: a) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh -2x + 3 > 0 v\u00e0 bi\u1ec3u di\u1ec5n tr\u00ean tr\u1ee5c s\u1ed1 t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3. b) T\u1eeb \u0111\u00f3 h\u00e3y ch\u1ec9 ra c\u00e1c kho\u1ea3ng m\u00e0 n\u1ebfu x l\u1ea5y gi\u00e1 tr\u1ecb trong \u0111\u00f3 th\u00ec nh\u1ecb th\u1ee9c f(x) = -2x + […]<\/p>\n","protected":false},"author":12,"featured_media":44340,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n