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{"id":21200,"date":"2018-03-13T23:44:24","date_gmt":"2018-03-13T16:44:24","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=21200"},"modified":"2018-03-13T23:45:36","modified_gmt":"2018-03-13T16:45:36","slug":"giai-tich-chuong-1-tap-bai-tap","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/","title":{"rendered":"Gi\u1ea3i t\u00edch – Ch\u01b0\u01a1ng 1 – \u00d4n t\u1eadp – B\u00e0i t\u1eadp"},"content":{"rendered":"

B\u00e0i 1 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0Ph\u00e1t bi\u1ec3u c\u00e1c \u0111i\u1ec1u ki\u1ec7n \u0111\u1ed3ng bi\u1ebfn v\u00e0 ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1. T\u00ecm c\u00e1c kho\u1ea3ng \u0111\u01a1n \u0111i\u1ec7u c\u1ee7a h\u00e0m s\u1ed1<\/strong><\/span><\/p>\n

y = -x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– x – 7;<\/p>\n

\"\"<\/p>\n

 <\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

– \u0110i\u1ec1u ki\u1ec7n \u0111\u1ed3ng bi\u1ebfn, ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1:<\/p>\n

Cho h\u00e0m s\u1ed1 y = f(x) x\u00e1c \u0111\u1ecbnh tr\u00ean K, h\u00e0m s\u1ed1 f(x):<\/p>\n

+ \u0110\u1ed3ng bi\u1ebfn (t\u0103ng) tr\u00ean K n\u1ebfu \u2200 x1<\/sub>, x2<\/sub>\u00a0\u2208 K: x1<\/sub>\u00a0< x2<\/sub>\u00a0=> f(x1<\/sub>) < f(x2<\/sub>).<\/p>\n

+ Ngh\u1ecbch bi\u1ebfn (gi\u1ea3m) tr\u00ean K \u2200 x1<\/sub>, x2<\/sub>\u00a0\u2208 K: x1<\/sub>\u00a0< x2<\/sub>\u00a0=> f(x1<\/sub>) > f(x2<\/sub>)<\/p>\n

– X\u00e9t h\u00e0m s\u1ed1 y = -x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– x – 7, ta c\u00f3:<\/p>\n

D = R<\/p>\n

y’ = -3x2<\/sup>\u00a0+ 4x – 1<\/p>\n

y’ = 0 => x = 1 ; x = 1\/3<\/p>\n

<\/div>\n

y’ > 0 v\u1edbi x \u2208 (1\/3; 1) v\u00e0 y’ < 0 v\u1edbi x \u2208 (-\u221e; 1\/3) \u222a (1; +\u221e)<\/p>\n

V\u1eady h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean (1\/3; 1) v\u00e0 ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; 1\/3) \u222a (1; +\u221e).<\/p>\n

L\u01b0u \u00fd:\u00a0B\u1ea1n n\u00ean k\u1ebb b\u1ea3ng bi\u1ebfn thi\u00ean \u0111\u1ec3 th\u1ea5y s\u1ef1 \u0111\u01a1n \u0111i\u1ec7u r\u00f5 r\u00e0ng h\u01a1n.<\/em><\/strong><\/p>\n

– X\u00e9t h\u00e0m s\u1ed1<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Ta c\u00f3: D = R \\ {1}<\/p>\n

\"\"<\/p>\n

=> H\u00e0m s\u1ed1 lu\u00f4n ngh\u1ecbch bi\u1ebfn tr\u00ean t\u1eebng kho\u1ea3ng (-\u221e; 1) v\u00e0 (1; +-\u221e)<\/p>\n

B\u00e0i 2 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0N\u00eau c\u00e1ch t\u00ecm c\u1ef1c \u0111\u1ea1i, c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1 nh\u1edd \u0111\u1ea1o h\u00e0m. T\u00ecm c\u00e1c c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1:<\/strong><\/span><\/p>\n

y = x4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 2<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

– C\u00e1ch t\u00ecm c\u1ef1c \u0111\u1ea1i, c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1 nh\u1edd \u0111\u1ea1o h\u00e0m:<\/p>\n

Quy t\u1eafc 1:<\/b><\/p>\n

1. T\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh.<\/p>\n

2. T\u00ednh f'(x). T\u00ecm c\u00e1c \u0111i\u1ec3m t\u1ea1i \u0111\u00f3 f'(x) b\u1eb1ng 0 ho\u1eb7c f'(x) kh\u00f4ng x\u00e1c \u0111\u1ecbnh.<\/p>\n

3. L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean.<\/p>\n

4. T\u1eeb b\u1ea3ng bi\u1ebfn thi\u00ean suy ra c\u00e1c \u0111i\u1ec3m c\u1ef1c tr\u1ecb.<\/p>\n

Quy t\u1eafc 2:<\/b><\/p>\n

1. T\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh.<\/p>\n

2. T\u00ednh f'(x). Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh f'(x) = 0 v\u00e0 k\u00ed hi\u1ec7u xi<\/sub>\u00a0(i = 1, 2, 3, …) l\u00e0 c\u00e1c nghi\u1ec7m c\u1ee7a n\u00f3.<\/p>\n

3. T\u00ednh f”(x) v\u00e0 f”(xi<\/sub>)<\/p>\n

4. N\u1ebfu f”(xi<\/sub>) > 0 th\u00ec xi<\/sub>\u00a0l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n

N\u1ebfu f”(xi<\/sub>) < 0 th\u00ec xi<\/sub>\u00a0l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n

– X\u00e9t h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 2, ta c\u00f3:<\/p>\n

y’ = 4x3<\/sup>\u00a0– 4x = 4x(x2<\/sup>\u00a0– 1)<\/p>\n

y’ = 0 \u21d4 4x(x2<\/sup>\u00a0– 1) = 0 => x = 0; x = \u00b11<\/p>\n

y” = 12x2<\/sup>\u00a0– 4<\/p>\n

D\u1ef1a v\u00e0o Quy t\u1eafc 2, ta c\u00f3:<\/p>\n

y”(0) = -4 < 0 => x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i.<\/p>\n

y”(-1) = y”(1) = 8 > 0 => x = \u00b11 l\u00e0 hai \u0111i\u1ec3m c\u1ef1c ti\u1ec3u.<\/p>\n

B\u00e0i 3 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0N\u00eau c\u00e1ch t\u00ecm ra ti\u1ec7m c\u1eadn ngang v\u00e0 ti\u1ec7m c\u1eadn d\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. \u00c1p d\u1ee5ng \u0111\u1ec3 t\u00ecm c\u00e1c ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1:<\/strong><\/span><\/p>\n

 <\/p>\n

\"\"<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

– C\u00e1ch t\u00ecm ti\u1ec7m c\u1eadn ngang:<\/p>\n

\u0110\u01b0\u1eddng th\u1eb3ng y = yo<\/sub>\u00a0l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = f(x) n\u1ebfu \u00edt nh\u1ea5t m\u1ed9t trong c\u00e1c \u0111i\u1ec1u ki\u1ec7n sau \u0111\u01b0\u1ee3c th\u1ecfa m\u00e3n<\/p>\n

 <\/p>\n

\"\"<\/p>\n

– C\u00e1ch t\u00ecm ti\u1ec7m c\u1eadn \u0111\u1ee9ng:<\/p>\n

\u0110\u01b0\u1eddng th\u1eb3ng x = xo<\/sub>\u00a0l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = f(x) n\u1ebfu \u00edt nh\u1ea5t m\u1ed9t trong c\u00e1c \u0111i\u1ec1u ki\u1ec7n sau \u0111\u01b0\u1ee3c th\u1ecfa m\u00e3n<\/p>\n

 <\/p>\n

\"\"<\/p>\n

– X\u00e9t h\u00e0m s\u1ed1<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

=> \u0110\u1ed3 th\u1ecb c\u00f3 ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 x = 2.<\/p>\n

\"\"<\/p>\n

 <\/p>\n

=> \u0110\u1ed3 th\u1ecb c\u00f3 ti\u1ec7m c\u1eadn ngang l\u00e0 y = -2.<\/p>\n

B\u00e0i 4 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0Nh\u1eafc l\u1ea1i s\u01a1 \u0111\u1ed3 kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1.<\/strong><\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

H\u00e0m s\u1ed1 y = f(x)<\/p>\n

C\u00e1c b\u01b0\u1edbc kh\u1ea3o s\u00e1t h\u00e0m s\u1ed1:<\/p>\n

1. T\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1<\/p>\n

2. S\u1ef1 bi\u1ebfn thi\u00ean<\/p>\n

– X\u00e9t chi\u1ec1u bi\u1ebfn thi\u00ean:<\/p>\n

+ T\u00ednh \u0111\u1ea1o h\u00e0m y’<\/p>\n

+ T\u00ecm c\u00e1c \u0111i\u1ec3m t\u1ea1i \u0111\u00f3 y’ b\u1eb1ng 0 ho\u1eb7c kh\u00f4ng x\u00e1c \u0111\u1ecbnh<\/p>\n

+ X\u00e9t d\u1ea5u c\u1ee7a \u0111\u1ea1o h\u00e0m y’ v\u00e0 suy ra chi\u1ec1u bi\u1ebfn thi\u00ean c\u1ee7a h\u00e0m s\u1ed1.<\/p>\n

– T\u00ecm c\u1ef1c tr\u1ecb<\/p>\n

– T\u00ecm c\u00e1c gi\u1edbi h\u1ea1n t\u1ea1i v\u00f4 c\u1ef1c, c\u00e1c gi\u1edbi h\u1ea1n v\u00f4 c\u1ef1c v\u00e0 t\u00ecm ti\u1ec7m c\u1eadn (n\u1ebfu c\u00f3)<\/p>\n

– L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean.<\/p>\n

3. V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1<\/p>\n

D\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean v\u00e0 c\u00e1c y\u1ebfu t\u1ed1 x\u00e1c \u0111\u1ecbnh \u1edf tr\u00ean \u0111\u1ec3 v\u1ebd \u0111\u1ed3 th\u1ecb.<\/p>\n

B\u00e0i 5 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0Cho h\u00e0m s\u1ed1 y = 2x2<\/sup>\u00a0+ 2mx + m – 1 c\u00f3 \u0111\u1ed3 th\u1ecb l\u00e0 (Cm<\/sub>), m l\u00e0 tham s\u1ed1.<\/strong><\/span><\/p>\n

a) Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 khi m = -1<\/strong><\/span><\/p>\n

b) X\u00e1c \u0111\u1ecbnh m \u0111\u1ec3 h\u00e0m s\u1ed1:<\/strong><\/span><\/p>\n

\u00a0\u00a0\u00a0 i) \u0110\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-1; +\u221e)<\/strong><\/span><\/p>\n

\u00a0\u00a0\u00a0 ii) C\u00f3 c\u1ef1c tr\u1ecb tr\u00ean kho\u1ea3ng (-1; +\u221e)<\/strong><\/span><\/p>\n

c) Ch\u1ee9ng minh r\u1eb1ng (Cm<\/sub>) lu\u00f4n c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t v\u1edbi m\u1ecdi m.<\/strong><\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0V\u1edbi m = -1 ta \u0111\u01b0\u1ee3c h\u00e0m s\u1ed1: y = 2x2<\/sup>\u00a0+ 2x<\/p>\n

– TX\u0110: D = R, h\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 ti\u1ec7m c\u1eadn.<\/p>\n

– S\u1ef1 bi\u1ebfn thi\u00ean:<\/p>\n

+ Chi\u1ec1u bi\u1ebfn thi\u00ean: y’ = 4x + 2<\/p>\n

y’ = 0 => x = -1\/2<\/p>\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; -1\/2), \u0111\u1ed3ng bi\u1ebfn tr\u00ean (-1\/2; +\u221e).<\/p>\n

+ C\u1ef1c tr\u1ecb: H\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u l\u00e0 (-1\/2; 3\/2)<\/p>\n

– \u0110\u1ed3 th\u1ecb:<\/p>\n

Ta c\u00f3: 2x2<\/sup>\u00a0+ 2x = 0 \u21d4 2x(x + 1) = 0<\/p>\n

=> x = 0; x = -1<\/p>\n

+ Giao v\u1edbi Ox: (0; 0); (-1; 0)<\/p>\n

+ Giao v\u1edbi Oy: (0; 0)<\/p>\n

\"\"<\/p>\n

b)<\/b>\u00a0X\u00e9t h\u00e0m s\u1ed1 y = 2x2<\/sup>\u00a0+ 2mx + m – 1<\/p>\n

y’ = 4x + 2m = 2(2x + m)<\/p>\n

y’ = 0 => x = -m\/2<\/p>\n

Ta c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u y’:<\/p>\n

\"\"<\/p>\n

=> h\u00e0m s\u1ed1 c\u00f3 c\u1ef1c tr\u1ecb t\u1ea1i x = -m\/2<\/p>\n

– H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-1; +\u221e)<\/p>\n

\"\"<\/p>\n

– H\u00e0m s\u1ed1 c\u00f3 c\u1ef1c tr\u1ecb tr\u00ean kho\u1ea3ng (-1; +\u221e) th\u00ec:<\/p>\n

\"\"<\/p>\n

c)<\/b>\u00a0Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb (Cm<\/sub>) v\u00e0 tr\u1ee5c Ox l\u00e0:<\/p>\n

2x2<\/sup>\u00a0+ 2mx + m – 1 = 0 \u00a0\u00a0\u00a0 (1)<\/p>\n

\u0394’ = m2<\/sup>\u00a0– 2(m – 1) = m2<\/sup>\u00a0– 2m + 2<\/p>\n

= (m + 1)2<\/sup>\u00a0+ 1 > 0 \u2200 m \u2208 R<\/p>\n

=> Ph\u01b0\u01a1ng tr\u00ecnh (1) lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t, ngh\u0129a l\u00e0 \u0111\u1ed3 th\u1ecb lu\u00f4n c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t v\u1edbi m\u1ecdi m (\u0111pcm).<\/p>\n

B\u00e0i 6 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0a) Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb (C) c\u1ee7a h\u00e0m s\u1ed1:<\/strong><\/span><\/p>\n

f(x) = -x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 9x + 2<\/strong><\/span><\/p>\n

b) Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh f'(x – 1) > 0.<\/strong><\/span><\/p>\n

c) Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u1ed3 th\u1ecb (C) t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 x0<\/sub>, bi\u1ebft r\u1eb1ng f'(x0<\/sub>) = -6.<\/strong><\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0Kh\u1ea3o s\u00e1t h\u00e0m s\u1ed1 f(x) = -x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 9x + 2<\/p>\n

– TX\u0110: D = R<\/p>\n

– S\u1ef1 bi\u1ebfn thi\u00ean:<\/p>\n

+ Chi\u1ec1u bi\u1ebfn thi\u00ean: f'(x) = -3x2<\/sup>\u00a0+ 6x + 9<\/p>\n

f'(x) = 0 \u21d4 -3x2<\/sup>\u00a0+ 6x + 9 = 0 \u21d4 x = -1; x = 3<\/p>\n

+ Gi\u1edbi h\u1ea1n:<\/p>\n

\"\"<\/p>\n

B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean (-1; 3) v\u00e0 ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; -1) v\u00e0 (3; +\u221e).<\/p>\n

+ C\u1ef1c tr\u1ecb:<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i (3; 29);<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i (-1; -3);<\/p>\n

– \u0110\u1ed3 th\u1ecb:<\/p>\n

\"\"<\/p>\n

b)<\/b>\u00a0Ta c\u00f3: f'(x – 1) > 0<\/p>\n

\u21d4 -3(x – 1)2<\/sup>\u00a0+ 6(x – 1) + 9 > 0<\/p>\n

\u21d4 -3(x2<\/sup>\u00a0– 2x + 1) + 6x – 6 + 9 > 0<\/p>\n

\u21d4 -3x2<\/sup>\u00a0+ 6x – 3 + 6x – 6 + 9 > 0<\/p>\n

\u21d4 -3x2<\/sup>\u00a0+ 12x > 0 \u21d4 -x2<\/sup>\u00a0+ 4x > 0<\/p>\n

\u21d4 x(4 – x) > 0 \u21d4 0 < x < 4<\/p>\n

c)<\/b>\u00a0Ta c\u00f3: f”(x) = -6x + 6<\/p>\n

Theo b\u00e0i: f”(xo<\/sub>) = -6 => -6xo<\/sub>\u00a0+ 6 = -6 => xo<\/sub>\u00a0= 2<\/p>\n

V\u1eady ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn v\u1edbi (C) t\u1ea1i \u0111i\u1ec3m xo<\/sub>\u00a0= 2 l\u00e0:<\/p>\n

y = f'(2)(x – 2) + f(2)<\/p>\n

y = (-3.22<\/sup>\u00a0+ 6.2 + 9)(x – 2) + (-23<\/sup>\u00a0+ 3.22<\/sup>\u00a0+ 9.2 + 2)<\/p>\n

y = 9(x – 2) + 24 =\u00a09x + 6<\/b><\/p>\n

B\u00e0i 7 (trang 45-46 SGK Gi\u1ea3i t\u00edch 12):\u00a0a) Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1:<\/span><\/strong><\/p>\n

y = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 1<\/p>\n

b) D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb (C), bi\u1ec7n lu\u1eadn s\u1ed1 nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh sau theo m:<\/p>\n

x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 1 = m\/2<\/p>\n

c) Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i v\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a \u0111\u1ed3 th\u1ecb (C).<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0Kh\u1ea3o s\u00e1t h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 1<\/p>\n

– TX\u0110: D = R<\/p>\n

– S\u1ef1 bi\u1ebfn thi\u00ean:<\/p>\n

+ Chi\u1ec1u bi\u1ebfn thi\u00ean: y’ = 3x2<\/sup>\u00a0+ 6x = 3x(x + 2)<\/p>\n

y’ = 0 \u21d4 x = 0 ; x = -2<\/p>\n

+ Gi\u1edbi h\u1ea1n:<\/p>\n

\"\"<\/p>\n

B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; -2) v\u00e0 (0; +\u221e).<\/p>\n

H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-2; 0).<\/p>\n

+ C\u1ef1c tr\u1ecb:<\/p>\n

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u l\u00e0 (0; 1).<\/p>\n

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i l\u00e0 (-2; 5).<\/p>\n

– \u0110\u1ed3 th\u1ecb:<\/p>\n

+ Giao v\u1edbi Oy: (0; 1).<\/p>\n

+ \u0110\u1ed3 th\u1ecb (C) \u0111i qua \u0111i\u1ec3m (\u20133; 1), (1; 5).<\/p>\n

\"\"<\/p>\n

b)<\/b>\u00a0S\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 1 = m\/2 b\u1eb1ng s\u1ed1 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb (C) v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = m\/2.<\/p>\n

\"\"<\/p>\n

(\u0110\u01b0\u1eddng th\u1eb3ng y = m\/2 l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c Ox c\u1eaft tr\u1ee5c Oy t\u1ea1i \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 b\u1eb1ng m\/2)<\/em><\/p>\n

C\u00e1ch l\u00e0m: D\u1ecbch chuy\u1ec3n song song \u0111\u01b0\u1eddng th\u1eb3ng (d) v\u1edbi tr\u1ee5c Ox t\u1eeb tr\u00ean xu\u1ed1ng d\u01b0\u1edbi (ho\u1eb7c t\u1eeb d\u01b0\u1edbi l\u00ean tr\u00ean) l\u00e0 d\u1ef1a v\u00e0o s\u1ed1 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (C) \u0111\u1ec3 bi\u1ec7n lu\u1eadn.<\/em><\/p>\n

Ngo\u00e0i ra, trong khi l\u00e0m b\u00e0i, b\u1ea1n kh\u00f4ng c\u1ea7n v\u1ebd l\u1ea1i h\u00ecnh, ch\u1ec9 c\u1ea7n v\u1ebd (d) l\u00ean tr\u00ean \u0111\u1ed3 th\u1ecb v\u1eeba v\u1ebd l\u00e0 \u0111\u01b0\u1ee3c.<\/em><\/p>\n

Bi\u1ec7n lu\u1eadn:<\/b>\u00a0T\u1eeb \u0111\u1ed3 th\u1ecb ta c\u00f3:<\/p>\n

+ m\/2 < 1 \u21d4 m < 2: ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 1 nghi\u1ec7m.<\/p>\n

+ m\/2 = 1 \u21d4 m = 2: Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m.<\/p>\n

+ 1 < m\/2 < 5 \u21d4 1 < m < 10: ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 3 nghi\u1ec7m.<\/p>\n

+ m\/2 > 5 \u21d4 m > 10: ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 1 nghi\u1ec7m s\u1ed1.<\/p>\n

V\u1eady:<\/p>\n

+ N\u1ebfu m < 2 ho\u1eb7c m > 10 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 1 nghi\u1ec7m duy nh\u1ea5t.<\/p>\n

+ N\u1ebfu 2 < m < 10 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 3 nghi\u1ec7m.<\/p>\n

+ N\u1ebfu m = 2 ho\u1eb7c m= 10 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m.<\/p>\n

c)<\/b>\u00a0\u0110i\u1ec3m c\u1ef1c \u0111\u1ea1i A(-2; 5) v\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u B(0; 1).<\/p>\n

\"\"<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng AB l\u00e0:<\/p>\n

2.(x – 0) + 1.(y – 1) = 0 (l\u1ea5y t\u1ecda \u0111\u1ed9 B)<\/p>\n

=> y = -2x + 1<\/p>\n

V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i v\u00e0 c\u1ef1c ti\u1ec3u l\u00e0:\u00a0y = -2x + 1<\/b><\/p>\n

B\u00e0i 8 (trang 46 SGK Gi\u1ea3i t\u00edch 12):\u00a0Cho h\u00e0m s\u1ed1:<\/strong><\/span><\/p>\n

f(x) = x3<\/sup>\u00a0– 3mx2<\/sup>\u00a0+ 3(2m – 1)x + 1 (m l\u00e0 tham s\u1ed1).<\/strong><\/span><\/p>\n

a) X\u00e1c \u0111\u1ecbnh m \u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean t\u1eadp x\u00e1c \u0111\u1ecbnh.<\/strong><\/span><\/p>\n

b) V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a tham s\u1ed1 m th\u00ec h\u00e0m s\u1ed1 c\u00f3 m\u1ed9t c\u1ef1c \u0111\u1ea1i v\u00e0 m\u1ed9t c\u1ef1c ti\u1ec3u<\/strong><\/span>?<\/p>\n

c) X\u00e1c \u0111\u1ecbnh m \u0111\u1ec3 f”(x) > 6x.<\/strong><\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0TX\u0110: D = R<\/p>\n

f'(x) = 3x2<\/sup>\u00a0– 6mx + 3(2m – 1)<\/p>\n

f'(x) = 0 \u21d4 3x2<\/sup>\u00a0– 6mx + 3(2m – 1) = 0 \u00a0\u00a0\u00a0 (1)<\/p>\n

\u0394’ = (-3m)2<\/sup>\u00a0– 3.3(2m – 1) = 9(m2<\/sup>\u00a0– 2m + 1)<\/p>\n

= 9(m – 1)2<\/sup><\/p>\n

\u0110\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean D th\u00ec f'(x) \u2265 0<\/p>\n

\u21d4 \u0394’ \u2264 0 \u21d4 9(m – 1)2<\/sup>\u00a0\u2264 0 =>\u00a0m = 1<\/b><\/p>\n

b)<\/b>\u00a0H\u00e0m s\u1ed1 c\u00f3 m\u1ed9t c\u1ef1c \u0111\u1ea1i v\u00e0 m\u1ed9t c\u1ef1c ti\u1ec3u khi v\u00e0 ch\u1ec9 khi ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t.<\/p>\n

\u21d4 \u0394’ > 0 \u21d4 9(m – 1)2<\/sup>\u00a0> 0 =>\u00a0m \u2260 1<\/b><\/p>\n

c)<\/b>\u00a0Ta c\u00f3: f”(x) = 6x – 6m<\/p>\n

f”(x) > 6x \u21d4 6x – 6m > 6x<\/p>\n

\u21d4 – 6m > 0 \u21d4\u00a0m < 0<\/b><\/p>\n

B\u00e0i 9 (trang 46 SGK Gi\u1ea3i t\u00edch 12):\u00a0a) Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb (C) c\u1ee7a h\u00e0m s\u1ed1:<\/strong><\/span><\/p>\n

\"\"<\/p>\n

b) Vi\u1ebft ph\u01b0\u01a1ng t\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u1ed3 th\u1ecb (C) t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh f”(x) = 0.<\/p>\n

c) Bi\u1ec7n lu\u1eadn theo tham s\u1ed1 m s\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: x4<\/sup>\u00a0– 6x2<\/sup>\u00a0+ 3 = m.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0Kh\u1ea3o s\u00e1t h\u00e0m s\u1ed1<\/p>\n

\"\"<\/p>\n

\u00a0TX\u0110: D = R<\/p>\n

– S\u1ef1 bi\u1ebfn thi\u00ean:<\/p>\n

+ Chi\u1ec1u bi\u1ebfn thi\u00ean: f'(x) = 2x3<\/sup>\u00a0– 6x = 2x(x2<\/sup>\u00a0– 3)<\/p>\n

f'(x) = 0 \u21d4 2x(x2<\/sup>\u00a0– 3) = 0 \u21d4 x = 0; x = \u00b1\u221a3<\/p>\n

+ Gi\u1edbi h\u1ea1n t\u1ea1i v\u00f4 c\u1ef1c:<\/p>\n

\"\"<\/p>\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean (-\u221a3; 0) v\u00e0 (\u221a3; +\u221e).<\/p>\n

H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; -\u221a3) v\u00e0 (0; \u221a3).<\/p>\n

+ C\u1ef1c tr\u1ecb:<\/p>\n

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i (0; 3\/2)<\/p>\n

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i (-\u221a3; -3) v\u00e0 (\u221a3; -3)<\/p>\n

– \u0110\u1ed3 th\u1ecb:<\/p>\n

\"\"<\/p>\n

b)<\/b>\u00a0Ta c\u00f3: f”(x) = 6x2<\/sup>\u00a0– 6 = 6(x2<\/sup>\u00a0– 1)<\/p>\n

f”(x) = 0 \u21d4 6(x2<\/sup>\u00a0– 1) \u21d4 x = \u00b11 => y = -1<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a (C) t\u1ea1i (-1; -1) l\u00e0:<\/p>\n

y = f'(-1)(x + 1) – 1 =>\u00a0y = 4x + 3<\/b><\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a (C) t\u1ea1i (1; -1) l\u00e0:<\/p>\n

y = f'(1)(x – 1) – 1 =>\u00a0y = -4x + 3<\/b><\/p>\n

c)<\/b>\u00a0Ta c\u00f3: x4<\/sup>\u00a0– 6x2<\/sup>\u00a0+ 3 = m<\/p>\n

\"\"<\/p>\n

S\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (*) ch\u00ednh b\u1eb1ng s\u1ed1 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb (C) v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = m\/2.<\/p>\n

Bi\u1ec7n lu\u1eadn:<\/b>\u00a0T\u1eeb \u0111\u1ed3 th\u1ecb:<\/p>\n

+ m\/2 < – 3 \u21d4 m < -6: ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m.<\/p>\n

+ m\/2 = -3 \u21d4 m = -6 : ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 ngi\u1ec7m.<\/p>\n

+ -3 < m\/2 < 3\/2 \u21d4 -6 < m < 3: ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 4 nghi\u1ec7m.<\/p>\n

+ m\/2 = 3\/2 \u21d4 m = 3: ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 3 nghi\u1ec7m.<\/p>\n

+ m\/2 > 3\/2 \u21d4 m > 3: ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m.<\/p>\n

V\u1eady:<\/p>\n

+) m < – 6 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m.<\/p>\n

+) m = – 6 ho\u1eb7c m > 3 th\u00ec PT c\u00f3 2 nghi\u1ec7m.<\/p>\n

+) m = 3 th\u00ec PT c\u00f3 3 nghi\u1ec7m.<\/p>\n

+) \u2013 6 < m < 3 th\u00ec PT c\u00f3 4 nghi\u1ec7m.<\/p>\n

<\/div>\n
\n

B\u00e0i 10 (trang 46 SGK Gi\u1ea3i t\u00edch 12):\u00a0Cho h\u00e0m s\u1ed1<\/strong><\/span><\/p>\n

y = -x4<\/sup>\u00a0+ 2mx2<\/sup>\u00a0– 2m + 1 (m tham s\u1ed1)<\/strong><\/span><\/p>\n

c\u00f3 \u0111\u1ed3 th\u1ecb l\u00e0 (Cm<\/sub>).<\/strong><\/span><\/p>\n

a) Bi\u1ec7n lu\u1eadn theo m s\u1ed1 c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1.<\/strong><\/span><\/p>\n

d) V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m th\u00ec (Cm<\/sub>) c\u1eaft tr\u1ee5c ho\u00e0nh?<\/strong><\/span><\/p>\n

c) X\u00e1c \u0111\u1ecbnh \u0111\u1ec3 (Cm<\/sub>) c\u00f3 c\u1ef1c \u0111\u1ea1i, c\u1ef1c ti\u1ec3u.<\/strong><\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0y’ = -4x3<\/sup>\u00a0+ 4mx = 4x(m – x2<\/sup>)<\/p>\n

y’ = 0 (1<\/b>) \u21d4 4x(m – x2<\/sup>) = 0 => x = 0; x2<\/sup>\u00a0= m<\/p>\n

– N\u1ebfu m \u2264 0 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 1 nghi\u1ec7m => h\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb.<\/p>\n

– N\u1ebfu m > 0 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 3 nghi\u1ec7m => h\u00e0m s\u1ed1 c\u00f3 3 c\u1ef1c tr\u1ecb.<\/p>\n

b)<\/b>\u00a0Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (Cm<\/sub>) v\u1edbi tr\u1ee5c ho\u00e0nh:<\/p>\n

-x4<\/sup>\u00a0+ 2mx2<\/sup>\u00a0– 2m + 1 = 0 (2)<\/p>\n

\u0110\u1eb7t x2<\/sup>\u00a0= t (t \u2265 0) khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (2) t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi:<\/p>\n

-t2<\/sup>\u00a0+ 2mt – 2m + 1 = 0 (3)<\/p>\n

(Cm<\/sub>) c\u1eaft tr\u1ee5c ho\u00e0nh khi ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 nghi\u1ec7m. \u0110i\u1ec1u n\u00e0y t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh (3) c\u00f3 nghi\u1ec7m kh\u00f4ng \u00e2m. C\u00f3 hai tr\u01b0\u1eddng h\u1ee3p:<\/p>\n

– TH1: Ph\u01b0\u01a1ng tr\u00ecnh (3) c\u00f3 2 nghi\u1ec7m tr\u00e1i d\u1ea5u:<\/p>\n

\"\"<\/p>\n

– TH2: Ph\u01b0\u01a1ng tr\u00ecnh (3) c\u00f3 2 nghi\u1ec7m \u0111\u1ec1u kh\u00f4ng \u00e2m:<\/p>\n

\"\"<\/p>\n

K\u1ebft h\u1ee3p TH1 v\u00e0 TH2 ta c\u00f3 v\u1edbi m\u1ecdi m th\u00ec \u0111\u1ed3 th\u1ecb (Cm<\/sub>) lu\u00f4n c\u1eaft tr\u1ee5c ho\u00e0nh.<\/p>\n

c)<\/b>\u00a0(Cm<\/sub>) c\u00f3 c\u1ef1c \u0111\u1ea1i, c\u1ef1c ti\u1ec3u khi ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t.<\/p>\n

\u21d4 x2<\/sup>\u00a0= m c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t<\/p>\n

\u21d4\u00a0m > 0<\/b><\/p>\n

B\u00e0i 11 (trang 46 SGK Gi\u1ea3i t\u00edch 12):\u00a0a) Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb (C) c\u1ee7a h\u00e0m s\u1ed1<\/strong><\/span><\/p>\n

\"\"<\/p>\n

b) Ch\u1ee9ng minh r\u1eb1ng v\u1edbi m\u1ecdi gi\u00e1 tr\u1ecb c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng y = 2x + m lu\u00f4n c\u1eaft (C) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t M v\u00e0 N.<\/p>\n

c) X\u00e1c \u0111\u1ecbnh m sao cho \u0111\u1ed9 d\u00e0i MN nh\u1ecf nh\u1ea5t.<\/p>\n

d) Ti\u1ebfp tuy\u1ebfn t\u1ea1i m\u1ed9t \u0111i\u1ec3m S b\u1ea5t k\u00ec c\u1ee7a C c\u1eaft hai ti\u1ec7m c\u1eadn c\u1ee7a C t\u1ea1i P v\u00e0 Q. Ch\u1ee9ng minh r\u1eb1ng S l\u00e0 trung \u0111i\u1ec3m c\u1ee7a PQ.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0Kh\u1ea3o s\u00e1t h\u00e0m s\u1ed1:<\/p>\n

– TX\u0110: D = R \\ (-1)<\/p>\n

– S\u1ef1 bi\u1ebfn thi\u00ean:<\/p>\n

+ Chi\u1ec1u bi\u1ebfn thi\u00ean:<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 lu\u00f4n ngh\u1ecbch bi\u1ebfn tr\u00ean D.<\/p>\n

+ C\u1ef1c tr\u1ecb: H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb.<\/p>\n

+ Ti\u1ec7m c\u1eadn:<\/p>\n

\"\"<\/p>\n

=> \u0110\u1ed3 th\u1ecb c\u00f3 ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 x = -1.<\/p>\n

\"\"<\/p>\n

=> \u0110\u1ed3 th\u1ecb c\u00f3 ti\u1ec7m c\u1eadn ngang l\u00e0 y = 1.<\/p>\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n

\"\"<\/p>\n

– \u0110\u1ed3 th\u1ecb:<\/p>\n

+ Giao v\u1edbi Ox: (-3; 0)<\/p>\n

+ Giao v\u1edbi Oy: (0; 3)<\/p>\n

\"\"<\/p>\n

b)<\/b>\u00a0Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (C) v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = 2x + m l\u00e0:<\/p>\n

\"\"<\/p>\n

D\u1ec5 th\u1ea5y x = -1 kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1).<\/p>\n

Ta c\u00f3: \u0394 = (m + 1)2<\/sup>\u00a0– 8(m – 3) = m2<\/sup>\u00a0– 6m + 25<\/p>\n

\u0394 = (m – 3)2<\/sup>\u00a0+ 16 > 0 \u2200 m<\/p>\n

=> Ph\u01b0\u01a1ng tr\u00ecnh (1) lu\u00f4n c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t kh\u00e1c -1.<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = 2x + m lu\u00f4n c\u1eaft (C) t\u1ea1i 2 \u0111i\u1ec3m ph\u00e2n bi\u1ec7t M v\u00e0 N.<\/p>\n

c)<\/b>\u00a0Gi\u1ea3 s\u1eed M(x1<\/sub>; y1<\/sub>), N(x2<\/sub>; y2<\/sub>) v\u1edbi x1<\/sub>, x2<\/sub>\u00a0l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00e0 y1<\/sub>\u00a0= 2x1<\/sub>\u00a0+ m, y2<\/sub>\u00a0= 2x2<\/sub>\u00a0+ m.<\/p>\n

\"\"<\/p>\n

MN nh\u1ecf nh\u1ea5t \u21d4 MN2<\/sup>\u00a0nh\u1ecf nh\u1ea5t b\u1eb1ng 20.<\/p>\n

D\u1ea5u “=” x\u1ea3y ra \u21d4 m – 3 = 0 \u21d4 m = 3<\/p>\n

Khi \u0111\u00f3 \u0111\u1ed9 d\u00e0i MN nh\u1ecf nh\u1ea5t = \u221a20 = 2\u221a5<\/p>\n

d)<\/b>\u00a0G\u1ecdi S(xo<\/sub>; yo<\/sub>) \u2208 (C).<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn (d) c\u1ee7a (C) t\u1ea1i S l\u00e0:<\/p>\n

\"\"<\/p>\n

– Giao \u0111i\u1ec3m c\u1ee7a (d) v\u1edbi ti\u1ec7m c\u1eadn \u0111\u1ee9ng x = -1 l\u00e0:<\/p>\n

\"\"<\/p>\n

– Giao \u0111i\u1ec3m c\u1ee7a (d) v\u1edbi ti\u1ec7m c\u1eadn ngang y = 1 l\u00e0: Q(2xo<\/sub>\u00a0+ 1; 1).<\/p>\n

– Trung \u0111i\u1ec3m c\u1ee7a PQ l\u00e0 I(x1<\/sub>; y1<\/sub>) c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0:<\/p>\n

\"\"<\/p>\n

Suy ra S(xo<\/sub>; yo<\/sub>) ch\u00ednh l\u00e0 trung \u0111i\u1ec3m c\u1ee7a PQ (\u0111pcm).<\/p>\n

\n

\u00a0B\u00e0i 12 (trang 47 SGK Gi\u1ea3i t\u00edch 12):\u00a0Cho h\u00e0m s\u1ed1<\/strong><\/span><\/p>\n

\"\"<\/p>\n

a) Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh f'(sin x) = 0.<\/p>\n

b) Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh f”(cos x) = 0.<\/p>\n

c) Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh f”(x) = 0.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a)<\/b>\u00a0Ta c\u00f3: f'(x) = x2<\/sup>\u00a0– x – 4<\/p>\n

=> f'(sinx) = 0 \u21d4 sin2<\/sup>x – sinx – 4 = 0<\/p>\n

V\u00ec sin2<\/sup>x \u2264 1; -sinx \u2265 1 \u2200x \u2208 R<\/p>\n

=> sin2<\/sup>x – sinx \u2264 2<\/p>\n

=> sin2<\/sup>x – sinx – 4 \u2264 -2 \u2200x \u2208 R<\/p>\n

Do \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh f'(sinx) = 0 v\u00f4 nghi\u1ec7m.<\/p>\n

b)<\/b>\u00a0Ta c\u00f3: f”(x) = 2x – 1<\/p>\n

=> f”(cosx) = 0 \u21d4 2cosx – 1 = 0<\/p>\n

\"\"<\/p>\n

c)<\/b>\u00a0f”(x) = 0<\/p>\n

\"\"<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 t\u1ea1i x = 1\/2 l\u00e0:<\/p>\n

\"\"<\/p>\n

 <\/p>\n

 <\/p>\n<\/div>\n

 <\/p>\n

 <\/p>\n<\/div>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 1 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0Ph\u00e1t bi\u1ec3u c\u00e1c \u0111i\u1ec1u ki\u1ec7n \u0111\u1ed3ng bi\u1ebfn v\u00e0 ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1. T\u00ecm c\u00e1c kho\u1ea3ng \u0111\u01a1n \u0111i\u1ec7u c\u1ee7a h\u00e0m s\u1ed1 y = -x3\u00a0+ 2×2\u00a0– x – 7;   L\u1eddi gi\u1ea3i: – \u0110i\u1ec1u ki\u1ec7n \u0111\u1ed3ng bi\u1ebfn, ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1: Cho h\u00e0m s\u1ed1 y = f(x) x\u00e1c […]<\/p>\n","protected":false},"author":3,"featured_media":21204,"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":"\nGi\u1ea3i t\u00edch - Ch\u01b0\u01a1ng 1 - \u00d4n t\u1eadp - B\u00e0i t\u1eadp<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Gi\u1ea3i t\u00edch - Ch\u01b0\u01a1ng 1 - \u00d4n t\u1eadp - B\u00e0i t\u1eadp\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 1 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0Ph\u00e1t bi\u1ec3u c\u00e1c \u0111i\u1ec1u ki\u1ec7n \u0111\u1ed3ng bi\u1ebfn v\u00e0 ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1. T\u00ecm c\u00e1c kho\u1ea3ng \u0111\u01a1n \u0111i\u1ec7u c\u1ee7a h\u00e0m s\u1ed1 y = -x3\u00a0+ 2x2\u00a0– x – 7;   L\u1eddi gi\u1ea3i: – \u0110i\u1ec1u ki\u1ec7n \u0111\u1ed3ng bi\u1ebfn, ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1: Cho h\u00e0m s\u1ed1 y = f(x) x\u00e1c […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/\" \/>\n<meta property=\"og:site_name\" content=\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam\" \/>\n<meta property=\"article:published_time\" content=\"2018-03-13T16:44:24+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2018-03-13T16:45:36+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/03\/bai-1-trang-45-sgk-giai-tich-12-1.png\" \/>\n\t<meta property=\"og:image:width\" content=\"109\" \/>\n\t<meta property=\"og:image:height\" content=\"74\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"H\u00e0 Trang\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"H\u00e0 Trang\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"18 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/\",\"url\":\"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/\",\"name\":\"Gi\u1ea3i t\u00edch - Ch\u01b0\u01a1ng 1 - \u00d4n t\u1eadp - B\u00e0i t\u1eadp\",\"isPartOf\":{\"@id\":\"https:\/\/lop12.edu.vn\/#website\"},\"datePublished\":\"2018-03-13T16:44:24+00:00\",\"dateModified\":\"2018-03-13T16:45:36+00:00\",\"author\":{\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\"},\"breadcrumb\":{\"@id\":\"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/lop12.edu.vn\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Gi\u1ea3i t\u00edch – Ch\u01b0\u01a1ng 1 – \u00d4n t\u1eadp – B\u00e0i t\u1eadp\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/lop12.edu.vn\/#website\",\"url\":\"https:\/\/lop12.edu.vn\/\",\"name\":\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12\",\"description\":\"\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/lop12.edu.vn\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\",\"name\":\"H\u00e0 Trang\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"caption\":\"H\u00e0 Trang\"},\"url\":\"https:\/\/lop12.edu.vn\/author\/trangvth2\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Gi\u1ea3i t\u00edch - Ch\u01b0\u01a1ng 1 - \u00d4n t\u1eadp - B\u00e0i t\u1eadp","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/lop12.edu.vn\/giai-tich-chuong-1-tap-bai-tap\/","og_locale":"en_US","og_type":"article","og_title":"Gi\u1ea3i t\u00edch - Ch\u01b0\u01a1ng 1 - \u00d4n t\u1eadp - B\u00e0i t\u1eadp","og_description":"B\u00e0i 1 (trang 45 SGK Gi\u1ea3i t\u00edch 12):\u00a0Ph\u00e1t bi\u1ec3u c\u00e1c \u0111i\u1ec1u ki\u1ec7n \u0111\u1ed3ng bi\u1ebfn v\u00e0 ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1. 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