B\u00e0i 16 (trang 22 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1: f(x) = sin^{4<\/sup>x+cos4<\/sup>\u2061x.<\/p>\n}

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

H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R.<\/p>\n

Ta c\u00f3 f(x) = (sin^{2<\/sup>\u2061\u2061x )2<\/sup>\u2061+(cos2<\/sup>\u2061\u2061x )2<\/sup>\u2061=(sin2<\/sup>\u2061\u2061x+cos2<\/sup>\u2061x )2<\/sup>\u2061-2sin2<\/sup>\u2061x.cos2<\/sup>\u2061x=1-1\/2 sin2<\/sup>\u2061\u20612x v\u1edbi x \u2208R.<\/p>\n}

f(x)\u22641,\u2200x \u2208R,f(0)=0.<\/p>\n

<\/p>\n

f(x)\u22651\/2,\u2200x \u2208R (do sin^{2<\/sup>2x\u22641);f(\u03c0\/4)=1-1\/2=1\/2.<\/p>\n}

<\/p>\n

B\u00e0i 17 (trang 22 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n

T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau:<\/p>\n

f(x)=x^{2<\/sup>+2x-5 tr\u00ean \u0111o\u1ea1n [-2; 3]<\/p>\n}

f(x)=x^{3<\/sup>\/3+2x2<\/sup>+3x-4 tr\u00ean [-4; 0]<\/p>\n}

f(x)=x+1\/x tr\u00ean kho\u1ea3ng (0; +\u221e)<\/p>\n

f(x)=-x^{2<\/sup>+2x+4 tr\u00ean \u0111o\u1ea1n [2; 4]<\/p>\n}

<\/p>\n

f(x)=x-1\/x tr\u00ean n\u1eeda kho\u1ea3ng (0; 2)<\/p>\n

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

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

<\/p>\n

<\/p>\n

C\u00e1ch 2, ta c\u00f3: f(-2) = -5; f(-1) = -6; f(3) = 10<\/p>\n

<\/p>\n

f’ (x)=x^{2<\/sup>+4x+3;f’ (x)=0 <=> x = -1 ho\u1eb7c x = -3<\/p>\n}

Ta c\u00f3: f(-4) = -16\/3;f(-3)=-4;f(-1)=-16\/3;f(0)=-4<\/p>\n

<\/p>\n

<\/p>\n

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

<\/p>\n

H\u00e0m s\u1ed1 kh\u00f4ng \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t tr\u00ean (0; +\u221e)<\/p>\n

f’ (x)=-2x+2;f'(x)=0 <=> x = 1 (lo\u1ea1i v\u00ec x = 1 kh\u00f4ng thu\u1ed9c [2;4])<\/p>\n

Ta c\u00f3: f(2)=4;f(4)=-4<\/p>\n

<\/p>\n

<\/p>\n

C\u00e1ch 1. B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n

<\/p>\n

<\/p>\n

C\u00e1ch 2. V\u00ec x \u2208 [0;1] n\u00ean Ph\u01b0\u01a1ng tr\u00ecnh f\u2019(x) = 0 v\u00f4 nghi\u1ec7m tr\u00ean [0; 1]<\/p>\n

Ta c\u00f3: f(0) = 2; f(1) = 11\/3<\/p>\n

<\/p>\n

f(x)=x-1\/x;f'(x)=1+(1\/x^{2<\/sup>) >0,\u2200x \u2208(0;2),f(x)li\u00ean t\u1ee5c tr\u00ean (0; 2] n\u00ean f(x) \u0111\u1ed3ng bi\u1ebfn tr\u00ean (0; 2]<\/p>\n}

<\/p>\n

H\u00e0m s\u1ed1 kh\u00f4ng \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t tr\u00ean n\u1eeda kho\u1ea3ng (0; 2].<\/p>\n

B\u00e0i 18 (trang 22 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau:<\/p>\n

a) y=2 sin^{2<\/sup>\u2061x+2sinx-1<\/p>\n}

b) y=cos^{2<\/sup>\u2061x-sinxcosx+4<\/p>\n}

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

\u0110\u1eb7t t = sin x, -1\u2264t\u22641<\/p>\n

y=f(t)=2t^{2<\/sup>+2t-1,t \u2208[-1;1]<\/p>\n}

Ta t\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a y = f(t) tr\u00ean [-1;1]. \u0110\u00f3 l\u00e0 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho tr\u00ean R<\/p>\n

f’ (t)=4t+2,f’ (t)=0 <=> t=-1\/2<\/p>\n

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

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

y=cos^{2<\/sup>x-sinxcosx+4=1-sin2<\/sup>2x-1\/2 sin2x+4<\/p>\n}

=-sin^{2<\/sup>\u20612x-1\/2 sin\u20612x+5<\/p>\n}

\u0110\u1eb7t t = sin 2x, -1\u2264t\u22641<\/p>\n

y=f(t)=-t^{2<\/sup>-1\/2 t+5;f’ (t)=-2t-1\/2;f’ (x)=0 <=> t=-1\/4<\/p>\n}

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

<\/p>\n

<\/p>\n

<\/p>\n

B\u00e0i 19 (trang 22 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

Cho tam gi\u00e1c \u0111\u1ec1u ABC c\u1ea1nh a. ng\u01b0\u1eddi ta d\u1ef1ng m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt MNPQ c\u00f3 c\u1ea1nh MN n\u1eb1m tr\u00ean BC, hai \u0111\u1ec9nh P v\u00e0 Q theo th\u1ee9 t\u1ef1 n\u1eb1m tr\u00ean hai c\u1ea1nh AC v\u00e0 AB c\u1ee7a tam gi\u00e1c. X\u00e1c \u0111\u1ecbnh v\u1ecb tr\u1ecb c\u1ee7a \u0111i\u1ec3m M sao cho h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t \u0111\u00f3 l\u00e0.<\/p>\n

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

<\/p>\n

\u0110\u1eb7t BM = x (0<x<a\/2)<\/p>\n

Ta c\u00f3: MN \u2013 a \u2013 2x; QM = BM.tan B =x \u221a3<\/p>\n

Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt MNPQ l\u00e0:<\/p>\n

S(x)=QM.MN=x \u221a3(a-2x)<\/p>\n

S(x)=\u221a3(ax-2x^{2<\/sup>)<\/p>\n}

B\u00e0i to\u00e1n tr\u1edf th\u00e0nh t\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a S(x) tr\u00ean kho\u1ea3ng (0; a\/2)<\/p>\n

Ta c\u00f3 S’ (x)=\u221a3 (a-4x);S’ (x)=0 <=> x=a\/4<\/p>\n

<\/p>\n

S \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t t\u1ea1i x=a\/4 v\u00e0 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u00e2t MNQP l\u00e0:<\/p>\n

<\/p>\n

B\u00e0i 20 (trang 22 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n

M\u1ed9t h\u1ee3p t\u00e1c x\u00e3 nu\u00f4i c\u00e1 trong h\u1ed3. n\u1ebfu tr\u00ean m\u1ed7i \u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t h\u1ed3 c\u00f3 n con c\u00e1 th\u00ec trung b\u00ecnh m\u1ed7i con c\u00e1 sau m\u1ed9t v\u1ee5 c\u00e2n n\u1eb7ng:<\/p>\n

P(n)=480-20n (gam)<\/p>\n

H\u1ecfi ph\u1ea3i th\u1ea3 bao nhi\u00eau c\u00e1 tr\u00ean m\u1ed9t \u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t h\u1ed3 \u0111\u1ec3 sau m\u1ed9t v\u1ee5 thu ho\u1ea1ch \u0111\u01b0\u1ee3c nhi\u1ec1u c\u00e1 nh\u1ea5t?<\/p>\n

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

N\u1ebfu tr\u00ean m\u1ed7i \u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t h\u1ed3 c\u00f3 n con c\u00e1 th\u00ec sau m\u1ed9t v\u1ee5, s\u1ed1 c\u00e1 tr\u00ean m\u1ed7i \u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch m\u1eb7t h\u1ed3 trung b\u00ecnh c\u00e2n n\u0103ng:<\/p>\n

f(n)=n-P(n)=480n-20n^{2<\/sup>\u00a0(gam)n \u2208N*<\/p>\n}

X\u00e9t h\u00e0m s\u1ed1 f(x) = 480x-20x^{2<\/sup>\u00a0tr\u00ean (0; +\u221e)<\/p>\n}

( Bi\u1ebfn n l\u1ea5y c\u00e1c gi\u00e1 tr\u1ecb nguy\u00ean d\u01b0\u01a1ng \u0111\u01b0\u1ee3c thay b\u1eb1ng bi\u1ebfn s\u1ed1 x l\u1ea5y c\u00e1c gi\u00e1 tr\u1ecb tr\u00ean kho\u1ea3ng (0; +\u221e))<\/p>\n

f’ (x)=480-40x;f’ (x)=0 <=>x = 12<\/p>\n

<\/p>\n

Tren (0; +\u221e) h\u00e0m s\u1ed1 f(x) \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t t\u1ea1i x = 12<\/p>\n

Suy ra tr\u00ean t\u1eadp h\u1ee3p N* c\u00e1c s\u1ed1 nguy\u00ean d\u01b0\u01a1ng, h\u00e0m s\u1ed1 f \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t t\u1ea1i \u0111i\u1ec3m n = 12.<\/p>\n

V\u1eady mu\u1ed1n thu ho\u1ea1ch \u0111\u01b0\u1ee3c nhi\u1ec1u c\u00e1 nh\u1ea5t sau m\u1ed9t v\u1ee5 th\u00ec tr\u00ean m\u1ed7i \u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t h\u1ed3 ph\u1ea3i th\u1ea3 12 con c\u00e1.<\/p>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 16 (trang 22 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):\u00a0 T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1: f(x) = sin4x+cos4\u2061x. L\u1eddi gi\u1ea3i: H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R. Ta c\u00f3 f(x) = (sin2\u2061\u2061x )2\u2061+(cos2\u2061\u2061x )2\u2061=(sin2\u2061\u2061x+cos2\u2061x )2\u2061-2sin2\u2061x.cos2\u2061x=1-1\/2 sin2\u2061\u20612x v\u1edbi x \u2208R. f(x)\u22641,\u2200x \u2208R,f(0)=0. f(x)\u22651\/2,\u2200x \u2208R (do sin22x\u22641);f(\u03c0\/4)=1-1\/2=1\/2. B\u00e0i 17 (trang […]<\/p>\n","protected":false},"author":3,"featured_media":23936,"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