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{"id":23922,"date":"2018-03-27T11:58:43","date_gmt":"2018-03-27T04:58:43","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=23922"},"modified":"2018-03-27T11:58:43","modified_gmt":"2018-03-27T04:58:43","slug":"dai-so-chuong-1-luyen-tap-trang-23-24","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/dai-so-chuong-1-luyen-tap-trang-23-24\/","title":{"rendered":"\u0110\u1ea1i s\u1ed1 – Ch\u01b0\u01a1ng 1 – Luy\u1ec7n t\u1eadp (trang 23-24)"},"content":{"rendered":"

B\u00e0i 21 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

T\u00ecm c\u1ef1c tr\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau:<\/p>\n

\"\"<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/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 = -1; fCT<\/sub>=f(-1)=-1\/2<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 1; fC\u0110<\/sub>=f(1)=1\/2<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

V\u1eady h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c tr\u1ecb t\u1ea1i \u0111i\u1ec3m x=-3\/2;f(-3\/2)=27\/4<\/p>\n

H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c \u0111\u1ea1i<\/p>\n

T\u1eadp x\u00e1c \u0111\u1ecbnh: [-5; 5]<\/p>\n

V\u1eady h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c tr\u1ecb t\u1ea1i \u0111i\u1ec3m x=-3\/2;f(-3\/2)=27\/4<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c tr\u1ecb t\u1ea1i \u0111i\u1ec3m x = 0; fC\u0110<\/sub>=f(0)=\u221a5<\/p>\n

H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c ti\u1ec3u.<\/p>\n

f(x) x\u00e1c \u0111\u1ecbnh tr\u00ean D = (-\u221e; -1] \u222a[1; +\u221e)<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

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

\n

B\u00e0i 22 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb \u0111\u1ec3 :<\/p>\n

\u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i v\u00e0 c\u1ef1c ti\u1ec3u.<\/p>\n

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

T\u1eadp x\u00e1c \u0111\u1ecbnh: D = R \\ {1}<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 nghi\u1ec7m x \u2260 1 khi v\u00e0 ch\u1ec9 khi m \u2260 1<\/p>\n

H\u00e0m s\u1ed1 c\u00f3 c\u1ef1c \u0111\u1ea1i v\u00e0 c\u1ef1c ti\u1ec3u khi v\u00e0 ch\u1ec9 khi Ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t kh\u00e1c 1, ngh\u0129a l\u00e0:<\/p>\n

V\u1eady f(x) \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i v\u00e0 c\u1ef1c ti\u1ec3u khi v\u00e0 ch\u1ec9 khi m >0<\/p>\n

B\u00e0i 23 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

\u0110\u1ed9 gi\u1ea3m huy\u1ebft \u00e1p c\u1ee7a m\u1ed9t b\u1ec7nh nh\u00e2n \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c G(x) = 0,025x2<\/sup>\u00a0(30-x). Trong \u0111\u00f3 x l\u00e0 li\u1ebfu l\u01b0\u1ee3ng thu\u1ed1c \u0111\u01b0\u1ee3c ti\u00eam cho b\u1ec7nh nh\u1eadn (x l\u1ea5y \u0111\u01a1n v\u1ecb l\u00e0 miligam). T\u00ednh li\u1ec1u l\u01b0\u1ee3ng thu\u1ed9c c\u1ea7n ti\u00eam cho b\u1ec7nh nh\u00e2n \u0111\u1ec3 huy\u1ebft \u00e1p gi\u1ea3m nhi\u1ec1u nh\u1ea5t v\u00e0 t\u00ednh \u0111\u1ed9 gi\u1ea3m \u0111\u00f3.<\/p>\n

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

Ta c\u00f3 G(x) = 0,75x2<\/sup>-0,025x2<\/sup>;G’ (x)=1,5x-0,075x2<\/sup>;G(x)’=0 <=> x = 0; x = 20<\/p>\n

V\u1eady li\u1ec1u l\u01b0\u1ee3ng c\u1ea7n ti\u00eam cho b\u1ec7nh nh\u00e2n huy\u1ebft \u00e1p gi\u1ea3m nhi\u1ec1u nh\u1ea5t l\u00e0 20mg. khi \u0111\u00f3, \u0111\u1ed9 gi\u1ea3m huy\u1ebft \u00e1p l\u00e0 100.<\/p>\n

B\u00e0i 24 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n

Cho parabol (P) y = x2<\/sup>\u00a0v\u00e0 \u0111i\u1ec3m A(-3; 0). X\u00e1c \u0111\u1ecbnh \u0111i\u1ec3m M thu\u1ed9c Parabol (P) sao cho kho\u1ea3ng c\u00e1ch AM ng\u1eafn nh\u1ea5t v\u00e0 t\u00ecm kho\u1ea3ng c\u00e1ch ng\u1eafn nh\u1ea5t \u0111\u00f3.<\/p>\n

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

G\u1ecdi M(x; x2<\/sup>) l\u00e0 m\u1ed9t \u0111i\u1ec3m b\u1ea5t k\u00ec tr\u00ean (P).<\/p>\n

Ta c\u00f3: AM2<\/sup>=(x-3)2<\/sup>+(x2<\/sup>\u00a0)2<\/sup>=x4<\/sup>+6x+9<\/p>\n

AM nh\u1ecf nh\u1ea5t <=> f(x) = x4<\/sup>+x2<\/sup>+6x+9 \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t.<\/p>\n

f’ (x)=4x2<\/sup>+2x+6=(x+1)(4x2<\/sup>-4x+6)<\/p>\n

f’ (x)=0 <=> x = -1<\/p>\n

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

f \u0111\u1ea1t nh\u1ecf nh\u1ea5t t\u1ea1i \u0111i\u1ec3m x = -1; f(-1) = 5. Suy ra, kho\u1ea3ng c\u00e1ch AM \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t khi M \u1edf v\u1ecb tr\u00ed \u0111i\u1ec3m Mo<\/sub>\u00a0(-1;1). L\u00fac \u0111\u00f3 AMo<\/sub>=\u221a5<\/p>\n

B\u00e0i 25 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

M\u1ed9t con c\u00e1 h\u1ed3i b\u01a1i ng\u01b0\u1ee3c d\u00f2ng \u0111\u1ec3 v\u01b0\u1ee3t qua m\u1ed9t kho\u1ea3ng c\u00e1ch l\u00e0 300 km. v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 6 km\/h. n\u1ebfu v\u1eadn t\u1ed1c b\u1edfi c\u1ee7a c\u00e1c khi n\u01b0\u1edbc \u0111\u1ee9ng y\u00ean l\u00e0 v (km\/h) th\u00ec n\u0103ng l\u01b0\u1ee3ng ti\u00eau hao c\u1ee7a c\u00e1 trong t gi\u1edd \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c E(v) = cv3<\/sup>t. Trong \u0111\u00f3 c l\u00e0 m\u1ed9t h\u1eb1ng s\u1ed1, E \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng Jun, t\u00ecm v\u1eadn t\u00f3c c\u1ee7a c\u00e1 khi d\u00f2ng n\u01b0\u1edbc \u0111\u1ee9ng y\u00ean \u0111\u1ec3 ti\u00eau hao n\u0103ng l\u01b0\u1ee3ng l\u00e0 \u00edt nh\u1ea5t.<\/p>\n

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

V\u1eadn t\u1ed1c c\u1ee7a c\u00e1c khi b\u1edfi ng\u01b0\u1ee3c d\u00f2ng n\u01b0\u1edbc l\u00e0 v \u2013 6 (km\/h). th\u1eddi gian c\u00e1 b\u01a1i \u0111\u1ec3 v\u01b0\u1ee3t qua m\u1ed9t kho\u1ea3ng c\u00e1ch 300 km l\u00e0:<\/p>\n

N\u0103ng l\u01b0\u1ee3ng ti\u00eau hao c\u1ee7a c\u00e1 \u0111\u1ec3 v\u01b0\u1ee3t kho\u1ea3ng c\u00e1ch \u0111\u00f3 l\u00e0:<\/p>\n

B\u00e0i to\u00e1n tr\u1edf th\u00e0nh t\u00ecm v > 6 \u0111\u1ec3 E(v) nh\u1ecf nh\u1ea5t?<\/p>\n

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

V\u1eady ti\u00eau hao \u00edt n\u0103ng l\u01b0\u1ee3ng nh\u1ea5t, c\u00e1 ph\u1ea3i b\u1edfi v\u1edbi v\u1eadn t\u1ed1c 9 km\/h (khi n\u01b0\u1edbc \u0111\u1ee9ng y\u00ean).<\/p>\n

B\u00e0i 26 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

Sau khi ph\u00e1t hi\u1ec7n m\u1ed9t b\u1ec7nh d\u1ecbch c\u00e1c chuy\u00ean gia \u00fd t\u00ea \u01b0\u1edbc t\u00ednh s\u1ed1 ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh k\u1ec3 t\u1eeb ng\u00e0y xu\u1ea5t hi\u1ec7n b\u1ec7nh nh\u00e2n \u0111\u1ea7u ti\u00ean \u0111\u1ebfn ng\u00e0y th\u1ee9 t l\u00e0:<\/p>\n

f(t)=45t2<\/sup>-t3<\/sup>;t=0;1;2;3\u2026;25.<\/p>\n

N\u1ebfu coi f l\u00e0 h\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean \u0111o\u1ea1n [0; 25] th\u00ec f\u2019(t) \u0111\u01b0\u1ee3c xem l\u00e0 t\u1ed1c \u0111\u1ecd truy\u1ec1n b\u1ec7nh (ng\u01b0\u1eddi\/ng\u00e0y) t\u1ea1i th\u1eddi \u0111i\u1ec3m t.<\/p>\n

T\u00ednh t\u1ed1c \u0111\u1ecd truy\u1ec1n b\u1ec7nh v\u00e0o ng\u00e0y th\u1ee9 5.<\/p>\n

X\u00e1c \u0111\u1ecbnh c\u00e1c ng\u00e0y m\u00e0 t\u1ed1c \u0111\u1ed9 truy\u1ec1n b\u1ec7nh l\u00e0 l\u1edbn nh\u1ea5t v\u00e0 t\u00ednh t\u1ed1c \u0111\u1ed9 \u0111\u00f3.<\/p>\n

X\u00e1c \u0111\u1ecbnh c\u00e1c ng\u00e0y m\u00e0 t\u1ed1c \u0111\u1ed9 truy\u1ec1n b\u1ec7nh l\u1edbn h\u1edbn 600.<\/p>\n

X\u00e9t chi\u1ec1u bi\u1ebfn thi\u00ean c\u1ee7a h\u00e0m s\u1ed1 f tr\u00ean \u0111o\u1ea1n [0; 25]<\/p>\n

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

S\u1ed1 ng\u01b0\u1eddi nhi\u1ec5m b\u1ec7nh k\u1ec3 t\u1eeb ng\u00e0y xu\u1ea5t hi\u1ec7n b\u1ec7nh nh\u00e2n \u0111\u1ea7u ti\u00ean \u0111\u1ebfn ng\u00e0y th\u1ee9 t l\u00e0 f(t) = 45t2<\/sup>-t3<\/sup>, t nguy\u00ean thu\u1ed9c \u0111o\u1ea1n [0; 25]<\/p>\n

\u0110\u1ec3 x\u00e9t t\u1ed1c \u0111\u1ed9 truy\u1ec1n b\u1ec7nh, xem f(t) l\u00e0 h\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean \u0111o\u1ea1n [0; 25]<\/p>\n

f’ (t)=90t-3t2<\/sup>=3t(30-t)<\/p>\n

T\u1ed1c \u0111\u1ed9 truy\u1ec1n b\u1ec7nh v\u00e0o ng\u00e0y th\u1ee9 5 l\u00e0: f\u2019(t) = 375 ng\u01b0\u1eddi\/ng\u00e0y<\/p>\n

B\u00e0i to\u00e1n tr\u1edd th\u00e0nh: t\u00ecm t \u2208[0;25] \u0111\u1ec3 f\u2019(t) l\u00e0 l\u1edbn nh\u1ea5t.<\/p>\n

Ta c\u00f3: f\u2019\u2019(t) = 90 \u2013 6t; f\u2019\u2019(t) = 0 <=> t = 15<\/p>\n

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

V\u1eady t\u1ed1c \u0111\u1ed9 truy\u1ec1n b\u1ec7nh l\u1edbn nh\u1ea5t l\u00e0 v\u00e0o ng\u00e0y th\u1ee9 15.<\/p>\n

T\u1ed1c \u0111\u1ed9 \u0111\u00f3 l\u00e0 f\u2019(15) = 675 ng\u01b0\u1eddi\/ng\u00e0y<\/p>\n

f\u2019(t) > 600 <=> 90t-3t2<\/sup>>600 <=>10<t<20<\/p>\n

t\u1eeb ng\u00e0y th\u1ee9 11 \u0111\u1ebfn ng\u00e0y th\u1ee9 19, t\u1ed1c \u0111\u1ed9 truy\u1ec1n b\u1ec7nh l\u00e0 l\u1edbn h\u01a1n 600 ng\u01b0\u1eddi m\u1ed7i ng\u00e0y.<\/p>\n

f\u2019(t) = 3t(30 \u2013 t) \u2200t \u2208(0;25); f(t) li\u00ean t\u1ee5c tr\u00ean [0; 25]<\/p>\n

=>f(t) \u0111\u1ed3ng bi\u1ebfn tr\u00ean [0; 25]<\/p>\n

\n

B\u00e0i 27 (trang 24 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 nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau:<\/p>\n

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

\u2200x \u2208(-3;1) m\u00e0 f(x) li\u00ean t\u1ee5c tr\u00ean [-3; 1] n\u00ean h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean [-3; 1]<\/p>\n

f\u2019(x) = 0 v\u00f4 nghi\u1ec7m (tr\u00ean [-3; 1])<\/p>\n

ta c\u00f3: f(1) = 1; f(-3) = 3 n\u00ean:<\/p>\n

\n
<\/div>\n<\/div>\n

b) T\u1eadp x\u00e1c \u0111\u1ecbnh: D = [-2; 2]<\/p>\n

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

C\u00e1ch 2.<\/strong> Ta c\u00f3: f(-2) = -2; f(\u221a2)=2\u221a2;f(2)=2<\/p>\n

So s\u00e1nh c\u00e1c gi\u00e1 tr\u1ecb tr\u00ean ta \u0111\u01b0\u1ee3c:<\/p>\n

f(x)=sin4<\/sup>x+1-sin2<\/sup>\u2061x+2=sin4<\/sup>x-sin2<\/sup>\u2061x+3<\/p>\n

\u0110\u1eb7t t = sin3<\/sup>\u2061x;t \u2208[0;1]. Khi \u0111\u00f3 ta c\u00f3 h(t) = t2<\/sup>-t+3,t \u2208[0;1]<\/p>\n

f’ (t)=2t-1;h’ (t)=0 <=> t=1\/2\u2208[0;1]<\/p>\n

h(0)=3;h(1\/2)=11\/4;h(1)=3<\/p>\n

\u0111\u1ea1t \u0111\u01b0\u1ee3c gi\u1edbi h\u1ea1n t\u1ea1i x = 0.<\/p>\n

\u0111\u1ea1t \u0111\u01b0\u1ee3c gi\u1edbi h\u1ea1n t\u1ea1i x = ?\/4<\/p>\n

f(x)=x-sin\u20612x tr\u00ean [-\u03c0\/2;\u03c0]<\/p>\n

Ta c\u00f3: f\u2019(x) = 1- 2cos2x; f\u2019(x) = 0<\/p>\n

So s\u00e1nh gi\u00e1 tr\u1ecb tr\u00ean ta \u0111\u01b0\u1ee3c:<\/p>\n

B\u00e0i 28 (trang 24 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n

Trong c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi l\u00e0 40 cm, h\u00e3y x\u00e1c \u0111\u1ecbnh c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch l\u1edbn nh\u1ea5t.<\/p>\n

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

C\u00e1ch 1.<\/strong> V\u00ec chu vi c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 40 cm n\u00ean k\u00edch th\u01b0\u1edbc l\u00e0 x, 20 -x (cm) (0 < x <20)<\/p>\n

Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1ead tr\u00ean l\u00e0:<\/p>\n

S(x) = x(20 \u2013 x) = 20x – x2<\/sup>\u00a0x \u2208(0;20)<\/p>\n

S’ (x)=20-x;S’ (x)=0 <=> x = 10<\/p>\n

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

V\u1eady trong t\u1ea5t c\u1ea3 c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 40 cm, h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh l\u00e0 10 cm c\u00f3 di\u1ec7n t\u00edch l\u1edbn nh\u1ea5t.<\/p>\n

C\u00e1ch 2.<\/strong> H\u01b0\u1edbng d\u1eabn: s\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u00f4 \u2013 si.<\/p>\n

Hai s\u1ed1 d\u01b0\u01a1ng a, b c\u00f3 t\u1ed5ng kh\u00f4ng \u0111\u1ed5i th\u00ec t\u00edch c\u1ee7a ch\u00fang l\u00e0 l\u1edbn nh\u1ea5t khi v\u00e0 ch\u1ec9 khi a = b<\/p>\n

Gi\u1ea3i<\/p>\n

G\u1ecdi a, b l\u00e0 hai c\u1ea1nh c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt, ta c\u00f3 a + b = 20 (a, b >0)<\/p>\n

V\u1eady h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh 10 cm l\u00e0 di\u1ec7n t\u00edch l\u1edbn nh\u1ea5t (trong t\u1ea5t c\u1ea3 c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 40 cm)<\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 21 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):\u00a0 T\u00ecm c\u1ef1c tr\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau: L\u1eddi gi\u1ea3i: B\u1ea3ng bi\u1ebfn thi\u00ean H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = -1; fCT=f(-1)=-1\/2 H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 1; fC\u0110=f(1)=1\/2 B\u1ea3ng bi\u1ebfn thi\u00ean: V\u1eady h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c tr\u1ecb t\u1ea1i \u0111i\u1ec3m x=-3\/2;f(-3\/2)=27\/4 […]<\/p>\n","protected":false},"author":3,"featured_media":23923,"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\u0110\u1ea1i s\u1ed1 - Ch\u01b0\u01a1ng 1 - Luy\u1ec7n t\u1eadp (trang 23-24)<\/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\/dai-so-chuong-1-luyen-tap-trang-23-24\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u0110\u1ea1i s\u1ed1 - Ch\u01b0\u01a1ng 1 - Luy\u1ec7n t\u1eadp (trang 23-24)\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 21 (trang 23 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):\u00a0 T\u00ecm c\u1ef1c tr\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 sau: L\u1eddi gi\u1ea3i: B\u1ea3ng bi\u1ebfn thi\u00ean H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = -1; 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