L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n m(x \u2013 4) = 5x \u2013 2 \u21d4(m – 5)x = 4m – 2<\/p>\n\n\n\n N\u1ebfu m – 5 \u2260 0 \u21d4 m \u2260 5 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t <\/p>\n\n\n\n x = (4m – 2)\/(m – 5)<\/p>\n\n\n\n N\u1ebfu m \u2013 5 = 0 \u21d4 m = 5, ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<\/p>\n\n\n\n 0.x = 18 \u21d2 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<\/p>\n\n\n\n V\u1eady v\u1edbi m \u2260 5 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t <\/p>\n\n\n\n x = (4m – 2)\/(m – 5)<\/p>\n\n\n\n V\u1edbi m = 5 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m.<\/p>\n\n\n\n Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 59<\/strong>: L\u1eadp b\u1ea3ng tr\u00ean v\u1edbi bi\u1ec7t th\u1ee9c thu g\u1ecdn \u0394\u2019.<\/ins><\/p>\n\n\n\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n B\u00e0i 1 (trang 62 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n <\/ins><\/p>\n\n\n\n B\u00e0i 2 (trang 62 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Gi\u1ea3i v\u00e0 bi\u1ec7n lu\u1eadn c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau theo tham s\u1ed1 m:<\/p>\n\n\n\n a) m(x – 2) = 3x + 1 ;<\/p>\n\n\n\n b) m2<\/sup>x + 6 = 4x + 3m ;<\/ins><\/p>\n\n\n\n c) (2m + 1)x – 2m = 3x – 2.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n a) m(x \u2013 2) = 3x + 1<\/p>\n\n\n\n \u21d4 mx \u2013 2m = 3x + 1<\/p>\n\n\n\n \u21d4 mx \u2013 3x = 1 + 2m<\/p>\n\n\n\n \u21d4 (m \u2013 3).x = 1 + 2m (1) <\/p>\n\n\n\n + X\u00e9t m \u2013 3 \u2260 0 \u21d4 m \u2260 3, ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m duy nh\u1ea5t \n<\/p>\n\n\n\n + X\u00e9t m \u2013 3 = 0 \u21d4 m = 3, pt (1) \u21d4 0x = 7. Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. <\/p>\n\n\n\n K\u1ebft lu\u1eadn: <\/p>\n\n\n\n + v\u1edbi m = 3, ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m <\/p>\n\n\n\n + v\u1edbi m \u2260 3, ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t \n<\/p>\n\n\n\n b) m2<\/sup>x + 6 = 4x + 3m<\/p>\n\n\n\n \u21d4 m2<\/sup>.x \u2013 4x = 3m \u2013 6<\/p>\n\n\n\n \u21d4 (m2<\/sup> \u2013 4).x = 3m \u2013 6 (2) <\/ins><\/p>\n\n\n\n + X\u00e9t m2<\/sup> \u2013 4 \u2260 0 \u21d4 m \u2260 \u00b12, ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 nghi\u1ec7m duy nh\u1ea5t: <\/p>\n\n\n\n + X\u00e9t m2<\/sup> \u2013 4 = 0 \u21d4 m = \u00b12<\/p>\n\n\n\n \u25cf V\u1edbi m = 2, pt (2) \u21d4 0x = 0 , ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m<\/p>\n\n\n\n \u25cf V\u1edbi m = \u20132, pt (2) \u21d4 0x = \u201312, ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. <\/p>\n\n\n\n K\u1ebft lu\u1eadn: <\/p>\n\n\n\n + m = 2, ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m<\/p>\n\n\n\n + m = \u20132, ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m <\/p>\n\n\n\n + m \u2260 \u00b12, ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t \n <\/p>\n\n\n\n c) (2m + 1)x \u2013 2m = 3x \u2013 2<\/p>\n\n\n\n \u21d4 (2m + 1)x \u2013 3x = 2m \u2013 2<\/p>\n\n\n\n \u21d4 (2m + 1 \u2013 3).x = 2m \u2013 2<\/p>\n\n\n\n \u21d4 (2m \u2013 2).x = 2m \u2013 2 (3) <\/p>\n\n\n\n + X\u00e9t 2m \u2013 2 \u2260 0 \u21d4 m \u2260 1, pt (3) c\u00f3 nghi\u1ec7m duy nh\u1ea5t \n<\/p>\n\n\n\n + X\u00e9t 2m \u2013 2 = 0 \u21d4 m = 1, pt (3) \u21d4 0.x = 0, ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m. <\/p>\n\n\n\n K\u1ebft lu\u1eadn : <\/p>\n\n\n\n + V\u1edbi m = 1, ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m<\/p>\n\n\n\n + V\u1edbi m \u2260 1, ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t x = 1. <\/p>\n\n\n\n Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n \u0110\u1ec3 gi\u1ea3i v\u00e0 bi\u1ec7n lu\u1eadn ph\u01b0\u01a1ng tr\u00ecnh quy \u0111\u01b0\u1ee3c v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t, ta c\u1ea7n : <\/p>\n\n\n\n + \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng a.x = b b\u1eb1ng c\u00e1ch chuy\u1ec3n h\u1ebft nh\u1eefng s\u1ed1 h\u1ea1ng\n ch\u1ee9a x v\u1ec1 b\u00ean tr\u00e1i, chuy\u1ec3n h\u1ebft nh\u1eefng s\u1ed1 h\u1ea1ng t\u1ef1 do v\u1ec1 b\u00ean ph\u1ea3i. <\/p>\n\n\n\n + X\u00e9t a \u2260 0, ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t x = b\/a<\/p>\n\n\n\n X\u00e9t a = 0, n\u1ebfu b = 0, pt c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m ; n\u1ebfu b \u2260 0, pt v\u00f4 nghi\u1ec7m. <\/p>\n\n\n\n + K\u1ebft lu\u1eadn.<\/p>\n\n\n\n B\u00e0i 3 (trang 62 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> C\u00f3 hai r\u1ed5\n qu\u00fdt ch\u1ee9a s\u1ed1 qu\u00fdt b\u1eb1ng nhau. N\u1ebfu l\u1ea5y 30 qu\u1ea3 \u1edf r\u1ed5 th\u1ee9 nh\u1ea5t \u0111\u01b0a sang r\u1ed5 \nth\u1ee9 hai th\u00ec s\u1ed1 qu\u1ea3 \u1edf r\u1ed5 th\u1ee9 hai b\u1eb1ng 1\/3 c\u1ee7a b\u00ecnh ph\u01b0\u01a1ng s\u1ed1 qu\u1ea3 c\u00f2n l\u1ea1i \u1edf\n r\u1ed5 th\u1ee9 nh\u1ea5t. H\u1ecfi s\u1ed1 qu\u1ea3 qu\u00fdt \u1edf m\u1ed7i r\u1ed5 l\u00fac ban \u0111\u1ea7u l\u00e0 bao nhi\u00eau?<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n G\u1ecdi s\u1ed1 qu\u00fdt ban \u0111\u1ea7u \u1edf m\u1ed7i r\u1ed5 l\u00e0 x (qu\u1ea3) <\/p>\n\n\n\n Mu\u1ed1n l\u1ea5y 30 qu\u1ea3 \u1edf r\u1ed5 th\u1ee9 nh\u1ea5t \u0111\u01b0a sang r\u1ed5 th\u1ee9 hai th\u00ec s\u1ed1 qu\u1ea3 \u1edf m\u1ed7i r\u1ed5 l\u00fac \u0111\u1ea7u ph\u1ea3i nhi\u1ec1u h\u01a1n 30 qu\u1ea3 hay x > 30.<\/p>\n\n\n\n Khi \u0111\u00f3 r\u1ed5 th\u1ee9 nh\u1ea5t c\u00f2n x \u2013 30 qu\u1ea3; r\u1ed5 th\u1ee9 hai c\u00f3 x + 30 qu\u1ea3. <\/p>\n\n\n\n V\u00ec s\u1ed1 qu\u1ea3 \u1edf r\u1ed5 th\u1ee9 hai b\u1eb1ng 1\/3 b\u00ecnh ph\u01b0\u01a1ng s\u1ed1 qu\u1ea3 c\u00f2n l\u1ea1i \u1edf r\u1ed5 th\u1ee9 nh\u1ea5t n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<\/ins><\/p>\n\n\n\n Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh (1):<\/p>\n\n\n\n V\u00ec x > 30 n\u00ean x = 45 th\u1ecfa m\u00e3n. <\/p>\n\n\n\n V\u1eady ban \u0111\u1ea7u m\u1ed7i r\u1ed5 c\u00f3 45 qu\u1ea3 cam. <\/p>\n\n\n\n Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n \u0110\u00e2y l\u00e0 d\u1ea1ng b\u00e0i gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 h\u1ecdc \u1edf l\u1edbp 8. <\/p>\n\n\n\n B\u01b0\u1edbc 1: L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh: <\/p>\n\n\n\n + Ch\u1ecdn \u1ea9n s\u1ed1 v\u00e0 \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n th\u00edch h\u1ee3p cho \u1ea9n s\u1ed1. <\/p>\n\n\n\n + Bi\u1ec3u di\u1ec5n c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng ch\u01b0a bi\u1ebft theo \u1ea9n v\u00e0 c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng \u0111\u00e3 bi\u1ebft; <\/p>\n\n\n\n + L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb m\u1ed1i quan h\u1ec7 gi\u1eefa c\u00e1c \u0111\u1ea1i l\u01b0\u01a1ng. <\/p>\n\n\n\n B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <\/p>\n\n\n\n B\u01b0\u1edbc 3: Ki\u1ec3m tra xem trong c\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh, nghi\u1ec7m n\u00e0o th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n, nghi\u1ec7m n\u00e0o kh\u00f4ng r\u1ed3i k\u1ebft lu\u1eadn. <\/p>\n\n\n\n B\u00e0i 4 (trang 62 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n\n\n\n a) 2x4<\/sup> – 7x2<\/sup> + 5 = 0 ; b) 3x4<\/sup> + 2x2<\/sup> – 1 = 0<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n a) 2x4<\/sup> \u2013 7x2<\/sup> + 5 = 0 (1) <\/p>\n\n\n\n T\u1eadp x\u00e1c \u0111\u1ecbnh: D = R. <\/p>\n\n\n\n \u0110\u1eb7t t = x2<\/sup>, \u0111i\u1ec1u ki\u1ec7n t \u2265 0. <\/p>\n\n\n\n Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh: <\/p>\n\n\n\n 2t2<\/sup> \u2013 7t + 5 = 0 <\/p>\n\n\n\n \u21d4 (2t \u2013 5) (t \u2013 1) = 0 <\/ins><\/p>\n\n\n\n b) 3x4<\/sup> + 2x2<\/sup> \u2013 1 = 0 (2) <\/p>\n\n\n\n T\u1eadp x\u00e1c \u0111\u1ecbnh : D = R. <\/p>\n\n\n\n \u0110\u1eb7t t = x2<\/sup>, \u0111i\u1ec1u ki\u1ec7n t \u2265 0 <\/p>\n\n\n\n Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (2) tr\u1edf th\u00e0nh : <\/p>\n\n\n\n 3t2<\/sup> + 2t \u2013 1 = 0 \u21d4 (3t \u2013 1)(t + 1) = 0 <\/p>\n\n\n\n B\u00e0i 5 (trang 62 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau b\u1eb1ng m\u00e1y t\u00ednh b\u1ecf t\u00fai (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba)<\/p>\n\n\n\n a) 2x2<\/sup> – 5x – 4 = 0 ; b) -3x2<\/sup> + 4x + 2 = 0<\/p>\n\n\n\n c) 3x2<\/sup> + 7x + 4 = 0 ; d) 9x2<\/sup> – 6x – 4 = 0.<\/ins><\/p>\n\n\n\n H\u01b0\u1edbng d\u1eabn c\u00e1ch gi\u1ea3i c\u00e2u a)<\/em>: N\u1ebfu s\u1eed d\u1ee5ng m\u00e1y t\u00ednh CASIO fx-500 MS, ta \u1ea5n li\u00ean ti\u1ebfp c\u00e1c ph\u00edm<\/p>\n\n\n\n m\u00e0n h\u00ecnh hi\u1ec7n ra x1<\/sub> = 3.137458609<\/p>\n\n\n\n \u1ea4n ti\u1ebfp \n m\u00e0n h\u00ecnh hi\u1ec7n ra x2<\/sub> = \u20130.637458608<\/p>\n\n\n\n L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba ta \u0111\u01b0\u1ee3c nghi\u1ec7m g\u1ea7n \u0111\u00fang c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 x1<\/sub> \u2248 3.137 v\u00e0 x2<\/sub> \u2248 \u20130.637.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i: S\u1eed d\u1ee5ng m\u00e1y t\u00ednh CASIO fx\u2013500 MS<\/p>\n\n\n\n <\/ins><\/p>\n\n\n\n * N\u1ebfu s\u1eed d\u1ee5ng c\u00e1c lo\u1ea1i m\u00e1y t\u00ednh CASIO fx \u2013 570, \u0111\u1ec3 v\u00e0o ch\u01b0\u01a1ng tr\u00ecnh gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc 2 c\u00e1c b\u1ea1n \u1ea5n nh\u01b0 sau: <\/p>\n\n\n\n r\u1ed3i sau \u0111\u00f3 nh\u1eadp c\u00e1c h\u1ec7 s\u1ed1 v\u00e0 \u0111\u01b0a ra k\u1ebft qu\u1ea3 nh\u01b0 CASIO fx\u2013500 MS tr\u00ean. <\/p>\n\n\n\n * N\u1ebfu s\u1eed d\u1ee5ng c\u00e1c lo\u1ea1i m\u00e1y t\u00ednh VINACAL, \u0111\u1ec3 v\u00e0o ch\u01b0\u01a1ng tr\u00ecnh gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc 2 c\u00e1c b\u1ea1n \u1ea5n nh\u01b0 sau: <\/p>\n\n\n\n r\u1ed3i sau \u0111\u00f3 nh\u1eadp c\u00e1c h\u1ec7 s\u1ed1 v\u00e0 \u0111\u01b0a ra k\u1ebft qu\u1ea3 nh\u01b0 tr\u00ean. <\/p>\n\n\n\n * C\u00e1c lo\u1ea1i m\u00e1y t\u00ednh CASIO fx\u2013570, VINACAL tr\u00ean khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \nv\u00f4 t\u1ef7 s\u1ebd cho nghi\u1ec7m ch\u00ednh x\u00e1c d\u01b0\u1edbi d\u1ea1ng c\u0103n th\u1ee9c, \u0111\u1ec3 nghi\u1ec7m hi\u1ec3n th\u1ecb \nd\u01b0\u1edbi d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n, c\u00e1c b\u1ea1n \u1ea5n n\u00fat \n<\/ins><\/p>\n\n\n\n V\u00ed d\u1ee5 \u0111\u1ec3 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean m\u00e1y t\u00ednh CASIO fx\u2013570 VN, c\u00e1c b\u1ea1n \u1ea5n nh\u01b0 sau: <\/p>\n\n\n\n B\u00e0i 6 (trang 62-63 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n\n\n\n a) |3x – 2| = 2x + 3 ;<\/p>\n\n\n\n b) |2x – 1| = |-5x – 2| ;<\/ins><\/p>\n\n\n\n d) |2x + 5| = x2<\/sup> + 5x + 1.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n a) |3x \u2013 2| = 2x + 3 (1)<\/p>\n\n\n\n T\u1eadp x\u00e1c \u0111\u1ecbnh: D = R. <\/p>\n\n\n\n + N\u1ebfu \n th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh 3x \u2013 2 = 2x + 3. T\u1eeb \u0111\u00f3 x = 5. <\/p>\n\n\n\n Gi\u00e1 tr\u1ecb x = 5 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n n\u00ean x = 5 l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (3). <\/p>\n\n\n\n + N\u1ebfu \n th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh 2 \u2013 3x = 2x + 3. T\u1eeb \u0111\u00f3 <\/p>\n\n\n\n Gi\u00e1 tr\u1ecb \n l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (3). <\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m x = 5 v\u00e0 \n<\/p>\n\n\n\n b) |2x – 1| = |-5x – 2| (2) <\/p>\n\n\n\n T\u1eadp x\u00e1c \u0111\u1ecbnh D = R. <\/p>\n\n\n\n Ta c\u00f3:<\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \n v\u00e0 x = \u20131.<\/ins><\/p>\n\n\n\n + X\u00e9t x > \u20131, khi \u0111\u00f3 x + 1 > 0 n\u00ean |x + 1| = x + 1. <\/p>\n\n\n\n Khi \u0111\u00f3 pt (3) <\/p>\n\n\n\n + X\u00e9t x < \u20131, khi \u0111\u00f3 x + 1 < 0 n\u00ean |x + 1| = \u2013x \u2013 1. <\/p>\n\n\n\n Khi \u0111\u00f3 pt (3) <\/p>\n\n\n\n (kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x < \u20131). <\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 \n<\/p>\n\n\n\n d) |2x + 5| = x2<\/sup> + 5x + 1 (4)<\/p>\n\n\n\n T\u1eadp x\u00e1c \u0111\u1ecbnh: D = R. <\/p>\n\n\n\n + X\u00e9t 2x + 5 \u2265 0 \u21d4 \n, khi \u0111\u00f3 |2x + 5| = 2x + 5<\/p>\n\n\n\n Khi \u0111\u00f3 pt (4) \u21d4 2x + 5 = x2<\/sup> + 5x + 1<\/p>\n\n\n\n \u21d4 x2<\/sup> + 3x \u2013 4 = 0<\/p>\n\n\n\n \u21d4 (x + 4)(x \u2013 1) = 0 <\/p>\n\n\n\n \u21d4 x = \u20134 (kh\u00f4ng th\u1ecfa m\u00e3n) ho\u1eb7c x = 1 (th\u1ecfa m\u00e3n)<\/p>\n\n\n\n + X\u00e9t 2x + 5 < 0 \u21d4 \n, khi \u0111\u00f3 |2x + 5| = \u20132x \u2013 5. <\/p>\n\n\n\n Khi \u0111\u00f3 pt (4) \u21d4 \u20132x \u2013 5 = x2<\/sup> + 5x + 1<\/p>\n\n\n\n \u21d4 x2<\/sup> + 7x + 6 = 0 <\/p>\n\n\n\n \u21d4 (x + 1)(x + 6) = 0 <\/p>\n\n\n\n \u21d4 x = \u20131 (kh\u00f4ng th\u1ecfa m\u00e3n) ho\u1eb7c x = \u20136 (th\u1ecfa m\u00e3n). <\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m x = 1 ho\u1eb7c x = \u20136.<\/p>\n\n\n\n Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n + \u0110\u1ec3 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ch\u1ee9a d\u1ea5u gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i ch\u00fang ta c\u1ea7n l\u00e0m\n m\u1ea5t d\u1ea5u gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i b\u1eb1ng c\u00e1ch chia tr\u01b0\u1eddng h\u1ee3p (tr\u01b0\u1eddng h\u1ee3p A(x) \u00e2m\n th\u00ec |A(x)| = \u2013A(x), tr\u01b0\u1eddng h\u1ee3p A(x) d\u01b0\u01a1ng th\u00ec |A(x)| = A(x)) ho\u1eb7c b\u00ecnh \nph\u01b0\u01a1ng c\u1ea3 hai v\u1ebf. <\/p>\n\n\n\n + \u1ede b\u01b0\u1edbc b\u00ecnh ph\u01b0\u01a1ng c\u1ea3 hai v\u1ebf, ta d\u00f9ng d\u1ea5u t\u01b0\u01a1ng \u0111\u01b0\u01a1ng khi bi\u1ebft r\u00f5 bi\u1ec3u th\u1ee9c \u1edf c\u1ea3 hai v\u1ebf c\u00f9ng \u00e2m ho\u1eb7c c\u00f9ng d\u01b0\u01a1ng. <\/p>\n\n\n\n Trong tr\u01b0\u1eddng h\u1ee3p ch\u01b0a bi\u1ebft d\u1ea5u c\u1ee7a m\u1ed9t trong hai v\u1ebf ho\u1eb7c c\u1ea3 hai v\u1ebf, ta ph\u1ea3i d\u00f9ng d\u1ea5u suy ra v\u00e0 th\u1eed l\u1ea1i nghi\u1ec7m. <\/p>\n\n\n\n + Ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng |f(x)| = |g(x)| khi gi\u1ea3i b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p ph\u00e1 d\u1ea5u gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i ta s\u1ebd c\u00f3 4 tr\u01b0\u1eddng h\u1ee3p: <\/p>\n\n\n\n \u25cf |f(x)| = g(x) \u21d4 f(x) = g(x) ho\u1eb7c \u2013f(x) = g(x) <\/p>\n\n\n\n \u25cf |f(x)| = \u2013 g(x) \u21d4 f(x) = \u2013g(x) ho\u1eb7c \u2013f(x) = \u2013g(x)<\/p>\n\n\n\n 4 tr\u01b0\u1eddng h\u1ee3p tr\u00ean ta c\u00f3 th\u1ec3 vi\u1ebft g\u1ecdn th\u00e0nh hai tr\u01b0\u1eddng h\u1ee3p f(x) = g(x) ho\u1eb7c f(x) = \u2013 g(x). <\/p>\n\n\n\n V\u1eady ta c\u00f3 |f(x)| = |g(x)| \u21d4 f(x) = g(x) ho\u1eb7c f(x) = \u2013 g(x). <\/p>\n\n\n\n B\u00e0i 7 (trang 63 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n a) \n (1) <\/p>\n\n\n\n \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: 5x + 6 \u2265 0 \u21d4 \n<\/p>\n\n\n\n T\u1eeb (1) \u21d2 5x + 6 = (x \u2013 6)2<\/sup> <\/p>\n\n\n\n \u21d4 5x + 6 = x2<\/sup> \u2013 12x + 36 <\/p>\n\n\n\n \u21d4 x2<\/sup> \u2013 17x + 30 = 0<\/p>\n\n\n\n \u21d4 (x \u2013 15)(x \u2013 2) = 0<\/p>\n\n\n\n \u21d4 x = 15 (th\u1ecfa m\u00e3n \u0110KX\u0110) ho\u1eb7c x = 2 (th\u1ecfa m\u00e3n \u0111kx\u0111). <\/p>\n\n\n\n Th\u1eed l\u1ea1i x = 15 l\u00e0 nghi\u1ec7m c\u1ee7a (1), x = 2 kh\u00f4ng ph\u1ea3i nghi\u1ec7m c\u1ee7a (1)<\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m x = 15. <\/p>\n\n\n\n b) \n (2)<\/p>\n\n\n\n \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: -2 \u2264 x \u2264 3<\/p>\n\n\n\n Ta c\u00f3 (2)<\/p>\n\n\n\n Th\u1eed l\u1ea1i th\u1ea5y x = 2 kh\u00f4ng ph\u1ea3i nghi\u1ec7m c\u1ee7a (2) <\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t x = \u20131<\/ins><\/p>\n\n\n\n c) \n (3) <\/p>\n\n\n\n T\u1eadp x\u00e1c \u0111\u1ecbnh: D = R. <\/p>\n\n\n\n T\u1eeb pt (3) \u21d2 2x2<\/sup> + 5 = (x + 2)2<\/sup> <\/p>\n\n\n\n \u21d4 2x2<\/sup> + 5 = x2<\/sup> + 4x + 4<\/p>\n\n\n\n \u21d4 x2<\/sup> \u2013 4x + 1 = 0<\/p>\n\n\n\n Th\u1eed l\u1ea1i th\u1ea5y ch\u1ec9 c\u00f3 x = 2 + \u221a3 l\u00e0 nghi\u1ec7m c\u1ee7a (3)<\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t x = 2 + \u221a3. <\/p>\n\n\n\n d) \n (4) <\/p>\n\n\n\n Ta c\u00f3 \n v\u1edbi m\u1ecdi x. <\/p>\n\n\n\n Do \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp x\u00e1c \u0111\u1ecbnh D = R. <\/p>\n\n\n\n T\u1eeb (4) \u21d2 4x2<\/sup> + 2x + 10 = (3x + 1)2<\/sup> <\/p>\n\n\n\n \u21d4 4x2<\/sup> + 2x + 10 = 9x2<\/sup> + 6x + 1<\/p>\n\n\n\n \u21d4 5x2<\/sup> + 4x \u2013 9 = 0<\/p>\n\n\n\n \u21d4 x = 1 ho\u1eb7c x = \u20139\/5<\/p>\n\n\n\n Th\u1eed l\u1ea1i th\u1ea5y ch\u1ec9 c\u00f3 x = 1 l\u00e0 nghi\u1ec7m c\u1ee7a (4) <\/p>\n\n\n\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t x = 1. <\/p>\n\n\n\n Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n + \u0110\u1ec3 gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n d\u01b0\u1edbi d\u1ea5u c\u0103n b\u1eadc hai, ta th\u01b0\u1eddng \nb\u00ecnh ph\u01b0\u01a1ng c\u1ea3 hai v\u1ebf \u0111\u1ec3 \u0111\u01b0a v\u1ec1 m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh kh\u00f4ng ch\u1ee9a \u1ea9n d\u01b0\u1edbi d\u1ea5u \nc\u0103n. <\/p>\n\n\n\n + Khi b\u00ecnh ph\u01b0\u01a1ng c\u1ea3 hai v\u1ebf c\u1ee7a m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh, ta d\u00f9ng d\u1ea5u t\u01b0\u01a1ng \n\u0111\u01b0\u01a1ng khi bi\u1ebft r\u00f5 bi\u1ec3u th\u1ee9c \u1edf c\u1ea3 hai v\u1ebf c\u00f9ng \u00e2m ho\u1eb7c c\u00f9ng d\u01b0\u01a1ng. <\/p>\n\n\n\n Trong tr\u01b0\u1eddng h\u1ee3p ch\u01b0a bi\u1ebft d\u1ea5u c\u1ee7a m\u1ed9t trong hai v\u1ebf ho\u1eb7c c\u1ea3 hai v\u1ebf, ta ph\u1ea3i d\u00f9ng d\u1ea5u suy ra v\u00e0 th\u1eed l\u1ea1i nghi\u1ec7m. \n<\/p>\n\n\n\n B\u00e0i 8 (trang 63 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2<\/sup> – 2(m + 1)x + 3m – 5 = 0<\/p>\n\n\n\n X\u00e1c \u0111\u1ecbnh m \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m g\u1ea5p ba nghi\u1ec7m kia. T\u00ednh c\u00e1c nghi\u1ec7m trong tr\u01b0\u1eddng h\u1ee3p \u0111\u00f3.<\/p>\n\n\n\n L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n Ta c\u00f3 : 3x2<\/sup> \u2013 2(m + 1)x + 3m \u2013 5 = 0 (1)<\/p>\n\n\n\n (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t khi \u0394\u2019 > 0 <\/p>\n\n\n\n \u21d4 (m + 1)2<\/sup> \u2013 3.(3m \u2013 5) > 0 <\/p>\n\n\n\n \u21d4 m2<\/sup> + 2m + 1 \u2013 9m + 15 > 0 <\/p>\n\n\n\n \u21d4 m2<\/sup> \u2013 7m + 16 > 0 <\/p>\n\n\n\n \u21d4 (m \u2013 7\/2)2<\/sup> + 15\/4 > 0 <\/p>\n\n\n\n \u0110i\u1ec1u n\u00e0y lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi m \u2208 R hay ph\u01b0\u01a1ng tr\u00ecnh (1) lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t., g\u1ecdi hai nghi\u1ec7m \u0111\u00f3 l\u00e0 x1<\/sub>; x2<\/sub><\/p>\n\n\n\n Khi \u0111\u00f3 theo \u0111\u1ecbnh l\u00fd Vi\u2013et ta c\u00f3 \n (I) <\/ins><\/p>\n\n\n\n Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m g\u1ea5p ba nghi\u1ec7m kia, gi\u1ea3 s\u1eed x2<\/sub> = 3.x1<\/sub>, khi thay v\u00e0o (I) suy ra : <\/p>\n\n\n\n * TH1 : m = 3, pt (1) tr\u1edf th\u00e0nh 3x2<\/sup> \u2013 8m + 4 = 0 c\u00f3 hai nghi\u1ec7m x1<\/sub> = 2\/3 v\u00e0 x2<\/sub> = 2 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n. <\/p>\n\n\n\n * TH2 : m = 7, pt (1) tr\u1edf th\u00e0nh 3x2<\/sup> \u2013 16m + 16 = 0 c\u00f3 hai nghi\u1ec7m x1<\/sub> = 4\/3 v\u00e0 x2<\/sub> = 4 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n. <\/p>\n\n\n\n K\u1ebft lu\u1eadn : m = 3 th\u00ec pt c\u00f3 hai nghi\u1ec7m l\u00e0 2\/3 v\u00e0 2. <\/p>\n\n\n\n m = 7 th\u00ec pt c\u00f3 hai nghi\u1ec7m 4\/3 v\u00e0 4. <\/p>\n\n\n\n Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n \u0110\u1ec3 gi\u1ea3i c\u00e1c b\u00e0i to\u00e1n t\u00ecm tham s\u1ed1 \u0111\u1ec3 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n n\u00e0o \u0111\u00f3 c\u1ea7n : <\/p>\n\n\n\n + T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a tham s\u1ed1 \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m ho\u1eb7c c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t. <\/p>\n\n\n\n + \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Vi\u2013et \u0111\u1ec3 bi\u1ec3u di\u1ec5n nghi\u1ec7m theo tham s\u1ed1 <\/p>\n\n\n\n + Bi\u1ebfn \u0111\u1ed5i c\u00e1c \u0111i\u1ec1u ki\u1ec7n v\u1ec1 nghi\u1ec7m theo \u0111\u1ec1 b\u00e0i \u0111\u1ec3 t\u00ecm ra tham s\u1ed1. <\/p>\n\n\n\n + Th\u1eed l\u1ea1i v\u00e0 k\u1ebft lu\u1eadn.\u2003<\/p>\n","protected":false},"excerpt":{"rendered":" Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 58: Gi\u1ea3i v\u00e0 bi\u1ec7n lu\u1eadn ph\u01b0\u01a1ng tr\u00ecnh sau theo tham s\u1ed1 m: m(x \u2013 4) = 5x \u2013 2. L\u1eddi gi\u1ea3i m(x \u2013 4) = 5x \u2013 2 \u21d4(m – 5)x = 4m – 2 N\u1ebfu m – 5 \u2260 0 \u21d4 m \u2260 […]<\/p>\n","protected":false},"author":12,"featured_media":44327,"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](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/tra-loi-cau-hoi-toan-10-dai-so-bai-2-trang-59.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-1-trang-62-sgk-dai-so-10.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-1-trang-62-sgk-dai-so-10-10.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-1-trang-62-sgk-dai-so-10-12.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-1-trang-62-sgk-dai-so-10-13.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-1-trang-62-sgk-dai-so-10-14.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-1-trang-62-sgk-dai-so-10-16.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-62-sgk-dai-so-10-3.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-62-sgk-dai-so-10-4.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-62-sgk-dai-so-10-5.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-62-sgk-dai-so-10-6.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-2-trang-62-sgk-dai-so-10-7.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-3-trang-62-sgk-dai-so-10-11.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-3-trang-62-sgk-dai-so-10-2.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-4-trang-62-sgk-dai-so-10-4.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-5-trang-62-sgk-dai-so-10-5.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-5-trang-62-sgk-dai-so-10-6.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-5-trang-62-sgk-dai-so-10-7.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-5-trang-62-sgk-dai-so-10-8.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-5-trang-62-sgk-dai-so-10-9.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-5-trang-62-sgk-dai-so-10-10.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-5-trang-62-sgk-dai-so-10-11.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-20.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-21.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-22.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-23.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-24.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-25.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-26.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-27.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-28.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-29.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-30.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-31.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-32.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-6-trang-62-sgk-dai-so-10-33.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-10.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-11.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-12.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-13.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-17.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-14.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-15.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-7-trang-63-sgk-dai-so-10-16.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-8-trang-63-sgk-dai-so-10-3.PNG\")
<\/figure>\n\n\n\n![\"Gi\u1ea3i](\"https:\/\/vietjack.com\/giai-toan-lop-10\/images\/bai-8-trang-63-sgk-dai-so-10-4.PNG\")
<\/figure>\n\n\n\n