V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: T = (-\u221e; -8) \u222a (-2; -4\/3) \u222a (1; 2)<\/p>\n\n\n\n

d)<\/strong> x^{2<\/sup> – x – 6 \u2264 0<\/p>\n\n\n\n}

X\u00e9t tam th\u1ee9c f(x) = x^{2<\/sup> – x – 6 c\u00f3 hai nghi\u1ec7m x = -2 v\u00e0 x = 3, h\u1ec7 s\u1ed1 a = 1 > 0 <\/p>\n\n\n\n}

Do \u0111\u00f3 f(x) \u2264 0 khi -2 \u2264 x \u2264 3. <\/p>\n\n\n\n

V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: T = [-2; 3]<\/p>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

Tam th\u1ee9c f(x) = ax^{2<\/sup> + bx + c c\u00f3 \u0394 = b2<\/sup> \u2013 4ac: <\/p>\n\n\n\n}

+ N\u1ebfu \u0394 < 0, f(x) c\u00f9ng d\u1ea5u v\u1edbi a v\u1edbi \u2200 x \u2208 R<\/p>\n\n\n\n

+ N\u1ebfu \u0394 = 0, f(x) c\u00f9ng d\u1ea5u v\u1edbi a v\u1edbi \u2200 x \u2260 \u2013b\/2a. <\/p>\n\n\n\n

+ N\u1ebfu \u0394 > 0, f(x) c\u00f9ng d\u1ea5u v\u1edbi a n\u1ebfu x < x1 ho\u1eb7c x > x^{2<\/sup>; <\/p>\n\n\n\n}

f(x) tr\u00e1i d\u1ea5u v\u1edbi a n\u1ebfu x_{1<\/sub> < x < x2<\/sub>; trong \u0111\u00f3 x1<\/sub>; x2<\/sub> l\u00e0 hai nghi\u1ec7m c\u1ee7a f(x) v\u00e0 x1<\/sub> < x2<\/sub>. <\/p>\n\n\n\n}

B\u00e0i 4 (trang 105 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau v\u00f4 nghi\u1ec7m<\/p>\n\n\n\n

a) (m – 2)x^{2<\/sup> + 2(2m – 3)x + 5m – 6 = 0<\/ins><\/p>\n\n\n\n}

b) (3 – m)x^{2<\/sup> – 2(m + 3)x + m + 2 = 0<\/p>\n\n\n\n}

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

a) (m – 2)x^{2<\/sup> + 2(2m – 3)x + 5m – 6 = 0 (1) <\/p>\n\n\n\n}

– N\u1ebfu m – 2 = 0 \u21d4 m = 2, khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh:<\/p>\n\n\n\n

2x + 4 = 0 \u21d4 x = -2 hay ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 m\u1ed9t nghi\u1ec7m<\/p>\n\n\n\n

Do \u0111\u00f3 m = 2 kh\u00f4ng ph\u1ea3i l\u00e0 gi\u00e1 tr\u1ecb c\u1ea7n t\u00ecm.<\/p>\n\n\n\n

– N\u1ebfu m – 2 \u2260 0 \u21d4 m \u2260 2 ta c\u00f3:<\/p>\n\n\n\n

\u0394’ = (2m – 3)^{2<\/sup> – (m – 2)(5m – 6)<\/p>\n\n\n\n}

= 4m^{2<\/sup> – 12m + 9 – 5m2<\/sup> + 6m + 10m – 12<\/p>\n\n\n\n}

= -m^{2<\/sup> + 4m – 3 = (-m + 3)(m – 1)<\/p>\n\n\n\n}

(1) v\u00f4 nghi\u1ec7m \u21d4 \u0394’ < 0 \u21d4 (-m + 3)(m – 1) < 0 \u21d4 m \u2208 (-\u221e; 1) \u222a (3; +\u221e)<\/ins><\/p>\n\n\n\n

V\u1eady v\u1edbi m \u2208 (-\u221e; 1) \u222a (3; +\u221e) th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m.<\/p>\n\n\n\n

b) (3 – m)x^{2<\/sup> – 2(m + 3)x + m + 2 = 0 (2)<\/p>\n\n\n\n}

– N\u1ebfu 3 – m = 0 \u21d4 m = 3 khi \u0111\u00f3 (2) tr\u1edf th\u00e0nh -6x + 5 = 0 \u21d4 x = 5\/6<\/p>\n\n\n\n

Do \u0111\u00f3 m = 3 kh\u00f4ng ph\u1ea3i l\u00e0 gi\u00e1 tr\u1ecb c\u1ea7n t\u00ecm.<\/p>\n\n\n\n

– N\u1ebfu 3 – m \u2260 0 \u21d4 m \u2260 3 ta c\u00f3:<\/p>\n\n\n\n

\u0394’ = (m + 3)^{2<\/sup> – (3 – m)(m + 2)<\/p>\n\n\n\n}

= m^{2<\/sup> + 6m + 9 – 3m – 6 + m2<\/sup> + 2m<\/p>\n\n\n\n}

= 2m^{2<\/sup> + 5m + 3 = (m + 1)(2m + 3)<\/p>\n\n\n\n}

(2) v\u00f4 nghi\u1ec7m \u21d4\u0394’ < 0\u21d4 (m + 1)(2m + 3) < 0 \u21d4 m \u2208 (-3\/2; -1)<\/p>\n\n\n\n

V\u1eady v\u1edbi m \u2208 (-3\/2; -1) th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m.<\/p>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

+ Ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng ax + b = 0 v\u00f4 nghi\u1ec7m khi a = 0 v\u00e0 b \u2260 0. <\/p>\n\n\n\n

+ Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai d\u1ea1ng ax^{2<\/sup> + bx + c = 0 v\u00f4 nghi\u1ec7m khi \u0394 = b2 \u2013 4ac < 0 ho\u1eb7c \u0394\u2019 = (b\/2)2<\/sup> \u2013 ac < 0. <\/p>\n","protected":false},"excerpt":{"rendered":"}

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 5 trang 100: 1) X\u00e9t tam th\u1ee9c b\u1eadc hai f(x) = x2 \u2013 5x + 4. T\u00ednh f(4), f(2), f(-1), f(0) v\u00e0 nh\u1eadn x\u00e9t v\u1ec1 d\u1ea5u c\u1ee7a ch\u00fang. 2) Quan s\u00e1t \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x2 \u2013 5x + 4 (h.32a)) v\u00e0 ch\u1ec9 ra […]<\/p>\n","protected":false},"author":12,"featured_media":44348,"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