C\u00e2u 8:<\/b>\u00a0Bi\u1ebft r\u1eb1ng log_{3<\/sub>y = (1\/2)log3<\/sub>u + log3<\/sub>v + 1. H\u00e3y bi\u1ec3u th\u1ecb y theo u v\u00e0 v<\/p>\n}

A. y = 3\u221auv \u00a0\u00a0\u00a0B. y = 3u^{2<\/sup>v \u00a0\u00a0\u00a0C. y = 3 + \u221au + v \u00a0\u00a0\u00a0D. y = (\u221auv)3<\/sup><\/p>\n}

C\u00e2u 9:<\/b>\u00a0T\u00ecm s\u1ed1 k sao cho 2^{x<\/sup>\u00a0= ekx<\/sup>\u00a0v\u1edbi m\u1ecdi s\u1ed1 th\u1ef1c x<\/p>\n}

A. k = \u221a2 \u00a0\u00a0\u00a0B. k = 2^{x<\/sup>\u00a0\u00a0\u00a0C. k = log_{2<\/sub>e \u00a0\u00a0\u00a0D. k = ln2<\/p>\n}}

C\u00e2u 10:<\/b>\u00a0\u0110\u1ed9 pH c\u1ee7a m\u1ed9t ch\u1ea5t \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi c\u00f4ng th\u1ee9c pH = -log[H^{+<\/sup>] trong \u0111\u00f3 H+<\/sup>\u00a0l\u00e0 n\u1ed3ng \u0111\u1ed9 ion hy\u0111r\u00f4 trong ch\u1ea5t \u0111\u00f3 t\u00ednh theo mol\/l\u00edt (mol\/L). X\u00e1c \u0111\u1ecbnh n\u1ed3ng \u0111\u1ed9 ion H+<\/sup>\u00a0c\u1ee7a m\u1ed9t ch\u1ea5t bi\u1ebft r\u1eb1ng \u0111\u1ed9 pH c\u1ee7a n\u00f3 l\u00e0 8,06<\/p>\n}

A. 8,7.10^{-9<\/sup>\u00a0mol\/L \u00a0\u00a0\u00a0B. 2,44.10-7<\/sup>\u00a0mol\/L<\/p>\n}

C. 2,74,4 mol\/L \u00a0\u00a0\u00a0D. 3,6.10^{-7<\/sup>\u00a0mol\/L<\/p>\n}

C\u00e2u 11:<\/b>\u00a0log125 b\u1eb1ng<\/p>\n

A. 5log3 \u00a0\u00a0\u00a0B. 3 – 3log2 \u00a0\u00a0\u00a0C. 100log1,25 \u00a0\u00a0\u00a0D. (log25)(log5)<\/p>\n

C\u00e2u 12:<\/b>\u00a0Cho a, b, c l\u00e0 c\u00e1c s\u1ed1 d\u01b0\u01a1ng. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c log_{a<\/sub>b^{2<\/sup>.logb<\/sub>c2<\/sup>.logc<\/sub>a2<\/sup><\/p>\n}}

A. 1\/8 \u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0C. 8\u00a0\u00a0\u00a0D. 6<\/p>\n

C\u00e2u 13:<\/b>\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c<\/p>\n

<\/p>\n

C\u00e2u 14:<\/b>\u00a0V\u1edbi 0 < x \u2260 1 , bi\u1ec3u th\u1ee9c<\/p>\n

<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n

8-A<\/td>\n

9-D<\/td>\n

10-A<\/td>\n

11-B<\/td>\n

12-C<\/td>\n

13-D<\/td>\n

14-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 8:<\/b><\/p>\n log3<\/sub>y = (1\/2)log3<\/sub>u + log3<\/sub>v + 1 <=> log3<\/sub>y = log3<\/sub>u1\/2<\/sup>\u00a0+ log3<\/sub>v + log3<\/sub>3 = log3<\/sub>(\u221au.v.3) => y = 3\u221au.v<\/p>\n C\u00e2u 9:<\/b><\/p>\n Ta c\u00f3: 2x<\/sup>\u00a0= (eln2<\/sup>)x = exln2<\/sup>\u00a0= ekx<\/sup>\u00a0=> k = ln2<\/p>\n C\u00e2u 10:<\/b><\/p>\n pH = -log[H+<\/sup>] => [H+<\/sup>] = 10-pH<\/sup>\u00a0= 10-8,06<\/sup>\u00a0\u2248 8,76.10-9<\/sup>(mol\/L)<\/p>\n C\u00e2u 11:<\/b><\/p>\n log125 = log(1000\/8) = log1000 – log8 = log103<\/sup>\u00a0– log23<\/sup>\u00a0= 3 – 3log2<\/p>\n C\u00e2u 12:<\/b><\/p>\n loga<\/sub>b2<\/sup>.logb<\/sub>c2<\/sup>.logc<\/sub>a2<\/sup>\u00a0= (2loga<\/sub>b)(2logb<\/sub>c)(2logc<\/sub>a) = 8loga<\/sub>b.logb<\/sub>c.logc<\/sub>a = 8loga<\/sub>c.logc<\/sub>a = 8<\/p>\n C\u00e2u 13:<\/b><\/p>\n <\/p>\n <\/p>\n C\u00e2u 14:<\/b><\/p>\n <\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 8:\u00a0Bi\u1ebft r\u1eb1ng log3y = (1\/2)log3u + log3v + 1. H\u00e3y bi\u1ec3u th\u1ecb y theo u v\u00e0 v A. y = 3\u221auv \u00a0\u00a0\u00a0B. y = 3u2v \u00a0\u00a0\u00a0C. y = 3 + \u221au + v \u00a0\u00a0\u00a0D. y = (\u221auv)3 C\u00e2u 9:\u00a0T\u00ecm s\u1ed1 k sao cho 2x\u00a0= ekx\u00a0v\u1edbi m\u1ecdi s\u1ed1 th\u1ef1c x A. k = \u221a2 \u00a0\u00a0\u00a0B. […]<\/p>\n","protected":false},"author":3,"featured_media":27519,"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