C\u00e2u 7:<\/b>\u00a0Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh log3<\/sub>36 – log3<\/sub>x > 1<\/p>\n A. 0 < x < 12 \u00a0\u00a0\u00a0B. x < 12\u00a0\u00a0\u00a0C. x > 12 \u00a0\u00a0\u00a0D. x < 1\/12<\/p>\n C\u00e2u 8:<\/b>\u00a0Cho<\/p>\n <\/p>\n T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a k<\/p>\n <\/p>\n C\u00e2u 9:<\/b>\u00a0Cho h\u00e0m s\u1ed1<\/p>\n <\/p>\n Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\n A. H\u00e0m s\u1ed1 c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb<\/p>\n B. H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 \u0111i\u1ec3m c\u1ef1c tr\u1ecb n\u00e0o<\/p>\n C. H\u00e0m s\u1ed1 c\u00f3 \u0111\u00fang m\u1ed9t \u0111i\u1ec3m c\u1ef1c tr\u1ecb v\u00e0 n\u00f3 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u<\/p>\n D. H\u00e0m s\u1ed1 c\u00f3 \u0111\u00fang m\u1ed9t \u0111i\u1ec3m c\u1ef1c tr\u1ecb v\u00e0 n\u00f3 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i<\/p>\n C\u00e2u 10:<\/b>\u00a0T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 y = \u221bx(8 – x) tr\u00ean \u0111o\u1ea1n [0; 8]<\/p>\n A. 4\u221b4\u00a0\u00a0\u00a0B. 5\u221b5 \u00a0\u00a0\u00a0C. 6\u221b2 \u00a0\u00a0\u00a0D. 10\u221b2<\/p>\n C\u00e2u 11:<\/b>\u00a0N\u1ebfu y = (log2<\/sub>3)(log3<\/sub>4)(log4<\/sub>5)…(log31<\/sub>32) th\u00ec<\/p>\n A. y = 5 \u00a0\u00a0\u00a0B. 4 < y < 5 \u00a0\u00a0\u00a0C. 5 < y < 6\u00a0\u00a0\u00a0D. y = 6<\/p>\n C\u00e2u 12:<\/b>\u00a0\u0110\u1eb7t log80 = a, log45 = b . H\u00e3y t\u00ednh log36 theo a v\u00e0 b<\/p>\n A. a + b – 1\u00a0\u00a0\u00a0B. b – a + 1 \u00a0\u00a0\u00a0C. a + b – 2\u00a0\u00a0\u00a0D. b – a + 2<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 7:<\/b><\/p>\n log3<\/sub>36 – log3<\/sub>x > log3<\/sub>36 – 1 > log3<\/sub>x <=> log3<\/sub>x < log3<\/sub>36 – log3<\/sub>3<\/p>\n <=> log3<\/sub>x < log3<\/sub>(36\/3) <=> log3<\/sub>x < log3<\/sub>12 <=> 0 < x < 12<\/p>\n C\u00e2u 8:<\/b><\/p>\n <\/p>\n C\u00e2u 9:<\/b><\/p>\n <\/p>\n Ta th\u1ea5y y\u2019 \u0111\u1ed5i d\u1ea5u t\u1eeb \u00e2m sang d\u01b0\u01a1ng khi \u0111i qua x = 6\/11 n\u00ean h\u00e0m s\u1ed1 c\u00f3 \u0111\u00fang m\u1ed9t \u0111i\u1ec3m c\u1ef1c tr\u1ecb x = 6\/11 v\u00e0 \u0111\u00f3 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u<\/p>\n C\u00e2u 10:<\/b><\/p>\n <\/p>\n Ta c\u00f3: y(0) = y(8) = 0, y(2) = 6\u221b2<\/p>\n <\/p>\n C\u00e2u 11:<\/b><\/p>\n S\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c loga<\/sub>blogb<\/sub>c = logc<\/sub>a ta t\u00ednh \u0111\u01b0\u1ee3c<\/p>\n y = [(log2<\/sub>3)(log3<\/sub>4)](log4<\/sub>5)…(log31<\/sub>32) = [log2<\/sub>4](log4<\/sub>5)…(log31<\/sub>32)<\/p>\n = [(log2<\/sub>4)(log4<\/sub>5)](log5<\/sub>6)…(log31<\/sub>32) = [log2<\/sub>3](log5<\/sub>6)…(log31<\/sub>32)<\/p>\n = … = log2<\/sub>32 = log2<\/sub>25<\/sup>\u00a0= 5<\/p>\n C\u00e1ch kh\u00e1c:<\/p>\n <\/p>\n C\u00e2u 12:<\/b><\/p>\n Ta c\u00f3: log36 = log(22<\/sup>.32<\/sup>) = 2(log2 + log3)<\/p>\n log80 = a => 3log2 + 1 = a => log2 = (a-1)\/3<\/p>\n log45 = b => b = log(32<\/sup>.5) = 2log3 + log5 = 2log3 + log(10\/2) = 2log3 + 1 – log2<\/p>\n => 2log3 = b – 1 + log2 = b – 1 + (a – 1)\/3 = b + (a\/3) – (4\/3)<\/p>\n T\u1eeb \u0111\u00f3:<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 7:\u00a0Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh log336 – log3x > 1 A. 0 < x < 12 \u00a0\u00a0\u00a0B. x < 12\u00a0\u00a0\u00a0C. x > 12 \u00a0\u00a0\u00a0D. x < 1\/12 C\u00e2u 8:\u00a0Cho T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a k C\u00e2u 9:\u00a0Cho h\u00e0m s\u1ed1 Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang? A. H\u00e0m s\u1ed1 c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb B. H\u00e0m […]<\/p>\n","protected":false},"author":3,"featured_media":27215,"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":"\n\n\n
\n 7-A<\/td>\n 8-D<\/td>\n 9-C<\/td>\n 10-C<\/td>\n 11-A<\/td>\n 12-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n