B\u00e0i 1 (trang 23-24 SGK Gi\u1ea3i t\u00edch 12):\u00a0T\u00ednh gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1:<\/strong><\/span><\/p>\n a) y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0– 9x + 35 tr\u00ean c\u00e1c \u0111o\u1ea1n [-4; 4] v\u00e0 [0; 5]<\/strong><\/span><\/p>\n b) y= x4<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 2 tr\u00ean c\u00e1c \u0111o\u1ea1n [0; 3] v\u00e0 [2; 5]<\/strong><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n a)<\/b>\u00a0TX\u0110: D = R.<\/p>\n y’ = 3x2<\/sup>\u00a0– 6x – 9; y’ = 0 => x = \u20131 ho\u1eb7c x = 3.<\/p>\n – X\u00e9t h\u00e0m s\u1ed1 tr\u00ean \u0111o\u1ea1n [-4; 4]<\/p>\n V\u00ec -1 v\u00e0 3 \u0111\u1ec1u thu\u1ed9c \u0111o\u1ea1n [-4; 4] n\u00ean ta t\u00ednh c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m t\u1ea1i c\u00e1c \u0111i\u1ec3m -4; 4; -1; 3.<\/p>\n Ta c\u00f3: y(-4) = -41; y(4)= 15; y(-1) = 40; y(3)= 8<\/p>\n V\u1eady, gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 tr\u00ean [-4; 4] l\u00e0:<\/p>\n <\/p>\n Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 tr\u00ean [-4; 4] l\u00e0:<\/p>\n <\/p>\n – Tr\u00ean \u0111o\u1ea1n [0; 5]: ta th\u1ea5y y’ = 0 t\u1ea1i x = 3 \u2208 [0; 5]<\/p>\n Ta c\u00f3: y(0) = 35; y(5)= 40; y(3)= 8<\/p>\n V\u1eady, gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 tr\u00ean [0; 5] l\u00e0:<\/p>\n <\/p>\n Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 tr\u00ean [0; 5] l\u00e0:<\/p>\n <\/p>\n (C\u00e1c ph\u1ea7n b, c, d) d\u01b0\u1edbi \u0111\u00e2y tr\u00ecnh b\u00e0y theo m\u1ed9t c\u00e1ch kh\u00e1c, ng\u1eafn g\u1ecdn h\u01a1n, nh\u01b0ng v\u1eabn b\u00e1m s\u00e1t theo c\u1ea5u tr\u00fac tr\u00ean.<\/p>\n b)<\/b>\u00a0TX\u0110: D = R<\/p>\n y’ = 4x3<\/sup>\u00a0– 6x<\/p>\n <\/p>\n c)<\/b>\u00a0TX\u0110: D = (-\u221e; 1) \u222a (1; +\u221e)<\/p>\n <\/p>\n => H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean D.<\/p>\n <\/p>\n d)<\/b>\u00a0TX\u0110: D = (-\u221e; 5\/4]<\/p>\n <\/p>\n => H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean D.<\/p>\n Khi \u0111\u00f3 tr\u00ean \u0111o\u1ea1n [-1; 1]:<\/p>\n <\/p>\n B\u00e0i 2 (trang 24 SGK Gi\u1ea3i t\u00edch 12):\u00a0Trong s\u1ed1 c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 c\u00f9ng chu vi 16cm, h\u00e3y t\u00ecm h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch l\u1edbn nh\u1ea5t.<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n N\u1eeda chu vi h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 16:2 = 8cm<\/p>\n G\u1ecdi chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 x (cm) th\u00ec c\u1ea1nh kia c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 (8 – x) (cm) (v\u1edbi x \u2208 [0; 8]).<\/p>\n Di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0:<\/p>\n y = S(x) = x(8 – x) = -x2<\/sup>\u00a0+ 8x<\/p>\n X\u00e9t h\u00e0m s\u1ed1 tr\u00ean ta c\u00f3: D = [0; 8]<\/p>\n y’= -2x + 8 = -2(x – 4)<\/p>\n y’ = 0 => x = 4<\/p>\n <\/p>\n H\u00e0m s\u1ed1 \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 4 (=> c\u1ea1nh c\u00f2n l\u1ea1i l\u00e0 8 – 4 = 4) hay trong s\u1ed1 c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 16cm th\u00ec h\u00ecnh vu\u00f4ng c\u00f3 di\u1ec7n t\u00edch l\u1edbn nh\u1ea5t.<\/p>\n (L\u01b0u \u00fd:<\/b>\u00a0Thay v\u00ec x\u00e9t max, min nh\u01b0 tr\u00ean, b\u1ea1n c\u0169ng c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng B\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4-si v\u1edbi hai s\u1ed1 x v\u00e0 x – 8 \u0111\u1ec3 suy ra k\u1ebft qu\u1ea3 t\u01b0\u01a1ng t\u1ef1.)<\/p>\n B\u00e0i 3 (trang 24 SGK Gi\u1ea3i t\u00edch 12):\u00a0Trong t\u1ea5t c\u1ea3 c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch 48 m2<\/sup>, h\u00e3y x\u00e1c \u0111\u1ecbnh h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi nh\u1ecf nh\u1ea5t.<\/strong><\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n G\u1ecdi \u0111\u1ed9 d\u00e0i m\u1ed9t c\u1ea1nh c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 x (m) th\u00ec \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u00f2n l\u1ea1i l\u00e0 48\/x (m) (\u0111i\u1ec1u ki\u1ec7n: x > 0).<\/p>\n Khi \u0111\u00f3 chu vi h\u00ecnh ch\u1eef nh\u1eadt l\u00e0:<\/p>\n <\/p>\n X\u00e9t h\u00e0m s\u1ed1 tr\u00ean (0; +\u221e):<\/p>\n <\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 4\u221a3 hay trong c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 c\u00f9ng di\u1ec7n t\u00edch 48 m2<\/sup>\u00a0th\u00ec h\u00ecnh vu\u00f4ng c\u1ea1nh 4\u221a3 m l\u00e0 h\u00ecnh c\u00f3 chu vi nh\u1ecf nh\u1ea5t.<\/p>\n B\u00e0i 4 (trang 24 SGK Gi\u1ea3i t\u00edch 12):\u00a0T\u00ednh gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau:<\/span><\/strong><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n a)<\/b> D = R<\/p>\n Ta th\u1ea5y: 1 + x2<\/sup>\u00a0\u2265 1<\/p>\n<\/div>\n <\/p>\n => H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t l\u00e0 4 khi 1 + x2<\/sup>\u00a0= 1 => x = 0<\/p>\n V\u1eady:<\/p>\n <\/p>\n (C\u00e1ch kh\u00e1c: t\u00ednh \u0111\u1ea1o h\u00e0m v\u00e0 l\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean)<\/p>\n b)<\/b> D = R<\/p>\n y’ = 12x2<\/sup>\u00a0– 12x3<\/sup>\u00a0= 12x2<\/sup>(x – 1)<\/p>\n y’ = 0 => x = 0 ; x = 1<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n<\/div>\n <\/p>\n T\u1eeb b\u1ea3ng bi\u1ebfn thi\u00ean suy ra: max y = y(1) = 1<\/p>\n <\/p>\n B\u00e0i 5 (trang 24 SGK Gi\u1ea3i t\u00edch 12):\u00a0T\u00ednh gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau:<\/strong><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n a)<\/b><\/p>\n – C\u00e1ch 1:<\/p>\n Ta c\u00f3: y = |x| \u2265 0 \u2200 x<\/p>\n => H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t l\u00e0 min y = 0 khi x = 0.<\/p>\n – C\u00e1ch 2:<\/p>\n <\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n T\u1eeb b\u1ea3ng bi\u1ebfn thi\u00ean suy ra: min y = 0<\/p>\n b)<\/b>\u00a0D = (0; +\u221e)<\/p>\n <\/p>\n y’ = 0 => x = 2 (lo\u1ea1i x = -2 v\u00ec \u2209 (0; +\u221e))<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n <\/p>\n T\u1eeb b\u1ea3ng bi\u1ebfn thi\u00ean suy ra: min y = y(2) = 4<\/p>\n B\u00e0i 1 (trang 23-24 SGK Gi\u1ea3i t\u00edch 12):\u00a0T\u00ednh gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1: a) y = x3\u00a0– 3×2\u00a0– 9x + 35 tr\u00ean c\u00e1c \u0111o\u1ea1n [-4; 4] v\u00e0 [0; 5] b) y= x4\u00a0– 3×2\u00a0+ 2 tr\u00ean c\u00e1c \u0111o\u1ea1n [0; 3] v\u00e0 [2; 5] L\u1eddi gi\u1ea3i: a)\u00a0TX\u0110: D = R. y’ = […]<\/p>\n","protected":false},"author":3,"featured_media":21473,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1298],"tags":[1377,1356,1355],"yoast_head":"\n