B\u00e0i 80 (trang 64 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n <\/p>\n A)\u0110\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-2; 3)<\/p>\n B)Ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-2; 3)<\/p>\n C)Ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -2)<\/p>\n D)\u0110\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-; +\u221e)<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n B\u00e0i 81 (trang 64 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n H\u00e0m s\u1ed1 f(x)=6x5<\/sup>-15x4<\/sup>+10x3<\/sup>-22<\/p>\n A)Ngh\u1ecbch bi\u1ebfn tr\u00ean R.<\/p>\n B)\u0110\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e;0) v\u00e0 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (0; +\u221e)<\/p>\n C)\u0110\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\n D)Ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (0; 1)<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n f’ (x)=30x2<\/sup>\u00a0(x-1)2<\/sup>\u22650 \u2200x \u2208R. Ch\u1ecdn C.<\/p>\n B\u00e0i 82 (trang 64 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n H\u00e0m s\u1ed1 y=sin\u2061x-x<\/p>\n A)\u0110\u1ed3ng bi\u1ebfn tr\u00ean R.<\/p>\n B)Ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e;0)<\/p>\n C)Ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e;0) v\u00e0 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (0; +\u221e)<\/p>\n D)Ngh\u1ecbch bi\u1ebfn tr\u00ean R.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n f’ (x)=cosx-1\u22640 \u2200x \u2208R. Ch\u1ecdn D.<\/p>\n B\u00e0i 83 (trang 64 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n H\u00e0m s\u1ed1 f(x)=x3<\/sup>-3x2<\/sup>-9x+11<\/p>\n A. Nh\u1eadn \u0111i\u1ec3m x = -1 l\u00e0m \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n B. Nh\u1eadn \u0111i\u1ec3m x = 3 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n C. Nh\u1eadn \u0111i\u1ec3m x = 1 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n D. Nh\u1eadn \u0111i\u1ec3m x = 3 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n f’ (x)=3(x2<\/sup>-2x-3)\u0111\u1ed5i d\u1ea5u t\u1eeb \u00e2m sang d\u01b0\u01a1ng t\u1ea1i \u0111i\u1ec3m x = 3. Ch\u1ecdn D.<\/p>\n B\u00e0i 84 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n H\u00e0m s\u1ed1 y=x4<\/sup>-4x3<\/sup>-5<\/p>\n A. Nh\u1eadn \u0111i\u1ec3m x = 3 l\u00e0m \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n B. Nh\u1eadn \u0111i\u1ec3m x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n C. Nh\u1eadn \u0111i\u1ec3m x = 3 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n D. Nh\u1eadn \u0111i\u1ec3m x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n y’=4x2<\/sup>\u00a0(x-3) \u0111\u1ed5i d\u1ea5u \u00e2m sang d\u01b0\u01a1ng t\u1ea1i \u0111i\u1ec3m x = 3. Ch\u1ecdn A.<\/p>\n B\u00e0i 85 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n S\u1ed1 \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y=x4<\/sup>-2x2<\/sup>-3 l\u00e0:<\/p>\n A.0<\/p>\n B. 1<\/p>\n C. 1<\/p>\n D. 2<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n V\u00ec f’ (x)=4x(x2<\/sup>-1)=0 c\u00f3 3 nghi\u1ec7m ph\u00e2n bi\u1ec7t v\u00e0 f\u2019(x) \u0111\u1ed5i d\u1ea5u qua c\u00e1c nghi\u1ec7m \u0111\u00f3. Ch\u1ecdn C.<\/p>\n B\u00e0i 86 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n S\u1ed1 \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 :<\/p>\n <\/p>\n A. 0<\/p>\n B. 2<\/p>\n C. 1<\/p>\n D. 3<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n y’=0 c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t x = -1 v\u00e0 x = 3 v\u00e0 y\u2019 \u0111\u1ed5i d\u1ea5u qua c\u00e1c \u0111i\u1ec3m \u0111\u00f3. Ch\u1ecdn B.<\/p>\n B\u00e0i 87 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n H\u00e0m s\u1ed1 f c\u00f3 \u0111\u1ea1o h\u00e0m f\u2019(x) = x2<\/sup>\u00a0(x+1)2<\/sup>(2x-1). S\u1ed1 \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0:<\/p>\n A. 1 \u00a0\u00a0\u00a0 B. 2 \u00a0\u00a0\u00a0 C. 1 \u00a0\u00a0\u00a0 D. 3<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n f’ (x)=1-2cos2x,f’ (-\u03c0\/6)=0 v\u00e0 \u0111\u1ed5i d\u1ea5u t\u1ea1i \u0111i\u1ec3m x=1\/2. Ch\u1ecdn A.<\/p>\n B\u00e0i 88 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n H\u00e0m s\u1ed1 y=x-sin\u20612 \u03c0+3<\/p>\n A)Nh\u1eadn \u0111i\u1ec3m x=-\u03c0\/6 l\u00e0m \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n B)Nh\u1eadn \u0111i\u1ec3m x=\u03c0\/2 l\u00e0m \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n C)Nh\u1eadn \u0111i\u1ec3m x=-\u03c0\/6 l\u00e0m \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n D)Nh\u1eadn \u0111i\u1ec3m x=-\u03c0\/2 l\u00e0m \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n f’ (x)=1-2cos2x,f’ (-\u03c0\/6)=0 v\u00e0 \u0111\u1ed5i d\u1ea5u t\u1eeb d\u01b0\u01a1ng sang \u00e2m t\u1ea1i \u0111i\u1ec3m x=-\u03c0\/6. Ch\u1ecdn C.<\/p>\n B\u00e0i 89 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 :<\/p>\n <\/p>\n A. -3 \u00a0\u00a0\u00a0 B. 1 \u00a0\u00a0\u00a0 C. -1 \u00a0\u00a0\u00a0 D. 0<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n B\u00e0i 90 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 y=3sinx-4cosx l\u00e0:<\/p>\n A. 3 \u00a0\u00a0\u00a0 B. -5 \u00a0\u00a0\u00a0 C. -4 \u00a0\u00a0\u00a0 D. -3<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n C\u00e1ch 1. (3sinx-4cosx)2<\/sup>\u2264(32<\/sup>+42<\/sup>\u00a0)(sin2<\/sup>\u2061x+cos2<\/sup>\u2061x )=25<\/p>\n <=>-5\u22643 sinx-4cosx\u226450. Ch\u1ecdn B.<\/p>\n C\u00e1ch 2. y=5 sin\u2061(x-\u03b1)v\u1edbi sin\u2061\u03b1=4\/5. Ch\u1ecdn B<\/p>\n B\u00e0i 91 (trang 65 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 f(x)=2x3<\/sup>\u00a03x2<\/sup>-12x+2 tr\u00ean \u0111o\u1ea1n [-1; 2] l\u00e0:<\/p>\n A. 6 \u00a0\u00a0\u00a0 B. 10 \u00a0\u00a0\u00a0 C. 15 \u00a0\u00a0\u00a0 D. 11<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n Ch\u1ecdn A.<\/p>\n B\u00e0i 92 (trang 66 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n Gi\u00e1 t\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 :<\/p>\n <\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n<\/div>\n <\/p>\n B\u00e0i 93 (trang 66 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n <\/p>\n A. \u0110\u01b0\u1eddng th\u1eb3ng x = -1 l\u00e0m ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a (C).<\/p>\n B. \u0110\u01b0\u1eddng th\u1eb3ng y = 2x \u2013 1 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a (C).<\/p>\n C. \u0110\u01b0\u1eddng th\u1eb3ng y = x + 1 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a (C).<\/p>\n D. \u0110\u01b0\u1eddng th\u1eb3ng y = x -2 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a (C).<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n Ch\u1ecdn D.<\/p>\n B\u00e0i 94 (trang 66 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n <\/p>\n A)\u0110\u01b0\u1eddng th\u1eb3ng x = 1 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a (C).<\/p>\n B)\u0110\u01b0\u1eddng th\u1eb3ng x = -1\/2 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a (C).<\/p>\n C)\u0110\u01b0\u1eddng th\u1eb3ng y = 1 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a (C).<\/p>\n D)\u0110\u01b0\u1eddng th\u1eb3ng y = -x + 1 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a (C).<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n 3+5x-2x2<\/sup>=0 c\u00f3 1 nghi\u1ec7m x=-1\/2 v\u00e0 1 nghi\u1ec7m x = 3.<\/p>\n => x=-1\/2 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a C). ch\u1ecdn A.<\/p>\n B\u00e0i 95 (trang 66 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n <\/p>\n A)\u0110\u01b0\u1eddng th\u1eb3ng x = 2 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a (C).<\/p>\n B)\u0110\u01b0\u1eddng th\u1eb3ng y= x – 1 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a (C).<\/p>\n C)\u0110\u01b0\u1eddng th\u1eb3ng y=-1\/5 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a (C).<\/p>\n D)\u0110\u01b0\u1eddng th\u1eb3ng y=-1\/2 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a (C).<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n Ch\u1ecdn C<\/p>\n B\u00e0i 96 (trang 66 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n <\/p>\n A. C\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng y = 1 t\u1ea1i hai \u0111i\u1ec3m.<\/p>\n B. C\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng y = 4 t\u1ea1i hai \u0111i\u1ec3m.<\/p>\n C. Ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng y = 0.<\/p>\n D. Kh\u00f4ng c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng y = -2.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n Ch\u1ecdn B<\/p>\n B\u00e0i 97 (trang 67 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n X\u00e9t Ph\u01b0\u01a1ng tr\u00ecnh: x3<\/sup>+3x2<\/sup>=m<\/p>\n A. V\u1edbi m = 5 , Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 3 nghi\u1ec7m.<\/p>\n B. V\u1edbi m = -1, Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 2 nghi\u1ec7m.<\/p>\n C. V\u1edbi m = 4, Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 3 nghi\u1ec7m ph\u00e2n bi\u1ec7t.<\/p>\n D. V\u1edbi m = 2 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 3 nghi\u1ec7m ph\u00e2n bi\u1ec7t.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n C\u00e1ch 1. Pt: x3<\/sup>+3x2<\/sup>=2 c\u00f3 3 nghi\u1ec7m ph\u00e2n bi\u1ec7t. ch\u1ecdn D.<\/p>\n C\u00e1ch 2. Minh h\u1ecda \u0111\u1ed3 th\u1ecb: yCT<\/sub>=0;yC\u0110<\/sub>=4. Ch\u1ecdn D.<\/p>\n B\u00e0i 98 (trang 67 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n <\/p>\n A.Nh\u1eadn \u0111i\u1ec3m (-1\/2;1\/2) l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n B.Nh\u1eadn \u0111i\u1ec3m (-1\/2;2) l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n C.Kh\u00f4ng c\u00f3 t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n D.Nh\u1eadn \u0111i\u1ec3m (1\/2;1\/2) l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n V\u00ec ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 x=-1\/2, ti\u1ec7m c\u1eadn ngang l\u00e0 y=1\/2 => nh\u1eadn \u0111i\u1ec3m (-1\/2;1\/2) l\u00e0m \u0111\u1ed1i x\u1ee9ng. ch\u1ecdn A.<\/p>\n<\/div>\n<\/div>\n B\u00e0i 99 (trang 67 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n S\u1ed1 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng cong y=x3<\/sup>-x2<\/sup>-2x+3 v\u00e0 y=x2<\/sup>-x+1 l\u00e0:<\/p>\n A. 0 \u00a0\u00a0\u00a0 B. 1 \u00a0\u00a0\u00a0 C. 3 \u00a0\u00a0\u00a0 D. 2<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n X\u00e9t Ph\u01b0\u01a1ng tr\u00ecnh: x3<\/sup>-x2<\/sup>-2x+3=x2<\/sup>-x+1<\/p>\n <=>x3<\/sup>-2x2<\/sup>-x+2= c\u00f3 3 nghi\u1ec7m ph\u00e2n bi\u1ec7t. ch\u1ecdn C.<\/p>\n B\u00e0i 100 (trang 67 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n C\u00e1c \u0111\u1ed3 th\u1ecb c\u1ee7a hai h\u00e0m s\u1ed1 y=3-1x v\u00e0 y=4x2<\/sup>\u00a0ti\u1ebfp x\u00fac v\u1edbi nhau t\u1ea1i M c\u00f3 ho\u00e0nh \u0111\u1ed9 l\u00e0:<\/p>\n A.X = -1 \u00a0\u00a0\u00a0 B. x = 1 \u00a0\u00a0\u00a0 C. x = 2 \u00a0\u00a0\u00a0 D. x=1\/2<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n V\u00ec y1<\/sub>‘ (1\/2)-y2<\/sub>‘ (1\/2)=> hai \u0111\u1ed3 th\u1ecb ti\u1ebfp x\u00fac nhau t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 x=1\/2. Ch\u1ecdn D.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":" B\u00e0i 80 (trang 64 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao): A)\u0110\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-2; 3) B)Ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-2; 3) C)Ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -2) D)\u0110\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-; +\u221e) L\u1eddi gi\u1ea3i: \u00a0f’ (x)=x2-x-6<0 \u2200x \u2208(-2;3). Ch\u1ecdn B. B\u00e0i 81 (trang 64 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao): H\u00e0m s\u1ed1 f(x)=6×5-15×4+10×3-22 […]<\/p>\n","protected":false},"author":3,"featured_media":23578,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1302],"tags":[1392,1393],"yoast_head":"\n