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{"id":27267,"date":"2018-04-17T12:35:55","date_gmt":"2018-04-17T05:35:55","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=27267"},"modified":"2018-04-17T12:36:50","modified_gmt":"2018-04-17T05:36:50","slug":"chuong-2-bai-tap-trac-nghiem-giai-tich-12-on-tap-chuong-2-phan-4","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-2-bai-tap-trac-nghiem-giai-tich-12-on-tap-chuong-2-phan-4\/","title":{"rendered":"Ch\u01b0\u01a1ng 2 – B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp ch\u01b0\u01a1ng 2 (Ph\u1ea7n 4)"},"content":{"rendered":"

C\u00e2u 25:<\/b>\u00a0T\u00ecm s\u1ed1 x kh\u00e1c 0 th\u1ecfa m\u00e3n (7x)14<\/sup>\u00a0= (14x)7<\/sup><\/p>\n

A. 7\u00a0\u00a0\u00a0B. 14\u00a0\u00a0\u00a0 C. 1\/7 \u00a0\u00a0\u00a0D. 2\/7<\/p>\n

C\u00e2u 26:<\/b>\u00a0T\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<\/p>\n

\"\"<\/p>\n

C\u00e2u 27:<\/b>\u00a0S\u1ed1 l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t \u0111\u00e0n chim sau th\u1eddi gian t th\u00e1ng k\u1ec3 t\u1eeb khi \u0111\u01b0\u1ee3c quan s\u00e1t \u0111\u01b0\u1ee3c \u01b0\u1edbc l\u01b0\u1ee3ng b\u1eb1ng c\u00f4ng th\u1ee9c<\/p>\n

\"\"<\/p>\n

Sau bao l\u00e2u k\u1ec3 t\u1eeb khi \u0111\u01b0\u1ee3c quan s\u00e1t th\u00ec \u0111\u00e0n chim c\u00f3 s\u1ed1 l\u01b0\u1ee3ng \u0111\u00f4ng nh\u1ea5t ?<\/p>\n

A. 1 th\u00e1ng \u00a0\u00a0\u00a0B. 4 th\u00e1ng\u00a0\u00a0\u00a0C. 5 th\u00e1ng \u00a0\u00a0\u00a0D. 8 th\u00e1ng<\/p>\n

C\u00e2u 28:<\/b>\u00a0T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb x th\u1ecfa m\u00e3n<\/p>\n

\"\"<\/p>\n

A. 2 \u00a0\u00a0\u00a0 B. 3 \u00a0\u00a0\u00a0C. 5 \u00a0\u00a0\u00a0D. 6<\/p>\n

C\u00e2u 29:<\/b>\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh 2x2\u00a0– 2x<\/sup>.3x<\/sup>\u00a0= 3\/2<\/p>\n

A. x = 1, x = 1 – log2<\/sub>3 \u00a0\u00a0\u00a0 C. x = 1, x = 1 + 2log2<\/sub>3<\/p>\n

B. x = 1, x = 1 + log2<\/sub>3 \u00a0\u00a0\u00a0 D. x = 1, x = 1 – 2log2<\/sub>3<\/p>\n

C\u00e2u 30:<\/b>\u00a0Cho ph\u01b0\u01a1ng tr\u00ecnh log5<\/sub>x + log3<\/sub>x = log5<\/sub>3.log9<\/sub>225 . Ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o sau \u0111\u00e2y kh\u00f4ng t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho?<\/p>\n

A. log5<\/sub>x + log3<\/sub>5.log5<\/sub>x = log5<\/sub>3.log3<\/sub>15<\/p>\n

B. log5<\/sub>x(1 + log3<\/sub>5) = log5<\/sub>3(1 + log3<\/sub>5)<\/p>\n

C. log5<\/sub>x = log3<\/sub>5<\/p>\n

D. log3<\/sub>x = 1<\/p>\n

C\u00e2u 31:<\/b>\u00a0Cho N > 1 . T\u00ecm s\u1ed1 th\u1ef1c x th\u1ecfa m\u00e3n<\/p>\n

\"\"<\/p>\n

\"B\u00e0i<\/p>\n

C\u00e2u 32:<\/b>\u00a0Cho a v\u00e0 b l\u00e0 hai s\u1ed1 th\u1ef1c th\u1ecfa m\u00e3n 3a<\/sup>\u00a0= 81b + 2<\/sup>\u00a0v\u00e0 125b<\/sup>\u00a0= 5a – 3<\/sup>\u00a0. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a ab<\/p>\n

A. -60 \u00a0\u00a0\u00a0B. -17\u00a0\u00a0\u00a0C. 12\u00a0\u00a0\u00a0D. 60<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
25-D<\/td>\n26-C<\/td>\n27-C<\/td>\n28-B<\/td>\n29-A<\/td>\n30-C<\/td>\n31-B<\/td>\n32-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 25:<\/b><\/p>\n

(7x)14<\/sup>\u00a0= (14x)7<\/sup>\u00a0=> 714<\/sup>.x14<\/sup>\u00a0= 147<\/sup>.x7<\/sup><\/p>\n

\"\"<\/p>\n

C\u00e2u 26:<\/b><\/p>\n

\"B\u00e0i<\/p>\n

<=> y’ = 0 <=> x2<\/sup>\u00a0= 4 <=> x = \u00b12. x = -2 \u2209 [-1; 4]<\/p>\n

Ta c\u00f3:<\/p>\n

\"\"<\/p>\n

y(4) = 4-2<\/sup>\u00a0(\u2248 0,54)<\/p>\n

\"\"<\/p>\n

C\u00e2u 27:<\/b><\/p>\n

\"\"<\/p>\n

P'(t) = 0 <=> t = 5.<\/p>\n

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

\"\"<\/p>\n

T\u1eeb \u0111\u00f3 ta th\u1ea5y sau 5 th\u00e1ng th\u00ec \u0111\u00e0n chim \u0111\u1ea1t s\u1ed1 l\u01b0\u1ee3ng \u0111\u00f4ng nh\u1ea5t<\/p>\n

C\u00e2u 28:<\/b><\/p>\n

\"\"<\/p>\n

\"B\u00e0i<\/p>\n

C\u00e2u 29:<\/b><\/p>\n

L\u1ea5y l\u00f4garit c\u01a1 s\u1ed1 2 hai v\u1ebf, ta \u0111\u01b0\u1ee3c:<\/p>\n

x2<\/sup>\u00a0– 2x + xlog2<\/sub>3 = log2<\/sub>(3\/2) <=> x2<\/sup>\u00a0– 2x + xlog2<\/sub>3 = log2<\/sub>3 – 1<\/p>\n

<=> x2<\/sup>\u00a0+ x(log2<\/sub>3 – 2) + 1 – log2<\/sub>3 = 0 <=> x = 1 ho\u1eb7c x = 1 – log2<\/sub>3<\/p>\n

C\u00e2u 30:<\/b><\/p>\n

T\u1eeb c\u00e1c ph\u01b0\u01a1ng \u00e1n \u0111\u00e3 cho, ta n\u00ean bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng ph\u01b0\u01a1ng tr\u00ecnh sao cho xu\u1ea5t hi\u1ec7n bi\u1ec3u th\u1ee9c log5<\/sub>x nh\u01b0 sau :<\/p>\n

log5<\/sub>x + log3<\/sub>x = log5<\/sub>3.log9<\/sub>225 <=> log5<\/sub>x + log3<\/sub>5.log5<\/sub>x = log5<\/sub>3.log22<\/sup><\/sub>152<\/sup><\/p>\n

<=> log5<\/sub>x + log3<\/sub>5.log5<\/sub>x = log5<\/sub>3.log3<\/sub>15 <=> log5<\/sub>x(1 + log3<\/sub>5) = log5<\/sub>3(1 + log3<\/sub>5)<\/p>\n

<=> log5<\/sub>x = log5<\/sub>3 <=> x = 3<\/p>\n

T\u1eeb \u0111\u00f3 ta th\u1ea5y ch\u1ec9 c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh log5<\/sub>x = log3<\/sub>5 l\u00e0 kh\u00f4ng t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho.<\/p>\n

Nh\u1eadn x\u00e9t. L\u01b0u \u00fd r\u1eb1ng hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 t\u01b0\u01a1ng \u0111\u01b0\u01a1ng n\u1ebfu ch\u00fang c\u00f3 c\u00f9ng t\u1eadp nghi\u1ec7m. Nh\u01b0 v\u1eady m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u1eddng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho th\u00ec kh\u00f4ng nh\u1ea5t thi\u1ebft ph\u1ea3i xu\u1ea5t hi\u1ec7n trong qu\u00e1 tr\u00ecnh gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho \u0111\u00f3.<\/p>\n

C\u00e2u 31:<\/b><\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi:<\/p>\n

logN<\/sub>x = logN<\/sub>2 + logN<\/sub>4 + logN<\/sub>6 + logN<\/sub>8 + logN<\/sub>10 = logN<\/sup>(2.4.6.8.10) = logN<\/sub>3840 => x = 3840<\/p>\n

C\u00e2u 32:<\/b><\/p>\n

T\u1eeb gi\u1ea3 thi\u1ebft c\u00f3: 3a<\/sup>\u00a0= 34(b + 2)<\/sup>\u00a0v\u00e0 53b<\/sup>\u00a0= 5a – 3<\/sup>.<\/p>\n

T\u1eeb \u0111\u00f3 suy ra: a = 4(b + 2) v\u00e0 3b = a – 3.<\/p>\n

gi\u1ea3i h\u1ec7 n\u00e0y t\u00ecm \u0111\u01b0\u1ee3c a = -12, b = -5. T\u1eeb \u0111\u00f3 ab = 60<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 25:\u00a0T\u00ecm s\u1ed1 x kh\u00e1c 0 th\u1ecfa m\u00e3n (7x)14\u00a0= (14x)7 A. 7\u00a0\u00a0\u00a0B. 14\u00a0\u00a0\u00a0 C. 1\/7 \u00a0\u00a0\u00a0D. 2\/7 C\u00e2u 26:\u00a0T\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 C\u00e2u 27:\u00a0S\u1ed1 l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t \u0111\u00e0n chim sau th\u1eddi gian t th\u00e1ng k\u1ec3 t\u1eeb khi \u0111\u01b0\u1ee3c quan s\u00e1t \u0111\u01b0\u1ee3c \u01b0\u1edbc l\u01b0\u1ee3ng b\u1eb1ng c\u00f4ng th\u1ee9c […]<\/p>\n","protected":false},"author":3,"featured_media":27268,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1447,1446],"yoast_head":"\nCh\u01b0\u01a1ng 2 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp ch\u01b0\u01a1ng 2 (Ph\u1ea7n 4)<\/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\/chuong-2-bai-tap-trac-nghiem-giai-tich-12-on-tap-chuong-2-phan-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 2 - 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