C\u00e2u 1:<\/b>\u00a0Bi\u1ebft 3 + 2log2<\/sub>x = log2<\/sub>y . H\u00e3y bi\u1ec3u th\u1ecb y theo x<\/p>\n A. y = 2x+3 \u00a0\u00a0\u00a0B. y = 8x2<\/sup> \u00a0\u00a0\u00a0C. y = x2<\/sup>+8 \u00a0\u00a0\u00a0D. y = 3x2<\/sup><\/p>\n C\u00e2u 2:<\/b>\u00a0N\u1ebfu x = (log8<\/sub>2)log2<\/sub>8<\/sup>\u00a0th\u00ec log3<\/sub>x b\u1eb1ng:<\/p>\n A. -3 \u00a0\u00a0\u00a0B. -1\/3\u00a0\u00a0\u00a0C. 1\/3 \u00a0\u00a0\u00a0D. 3<\/p>\n C\u00e2u 3:<\/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>] l\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 2,44<\/p>\n A. 1,1.108 mol\/L \u00a0\u00a0\u00a0 C. 3,6.10-3 mol\/L<\/p>\n B. 3,2.10-4 mol\/L \u00a0\u00a0\u00a0D. 3,7.10-3 mol\/L<\/p>\n C\u00e2u 4:<\/b>\u00a0R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<\/p>\n <\/p>\n C\u00e2u 5:<\/b>\u00a0V\u1edbi m\u1ed7i s\u1ed1 nguy\u00ean n, n > 1, \u0111\u1eb7t an<\/sub>\u00a0= (logn<\/sub>2002)-1<\/sup>, b = a2<\/sub>\u00a0+ a3<\/sub>\u00a0+ a4<\/sub>\u00a0+ a5<\/sub>\u00a0v\u00e0 c = a10<\/sub>\u00a0+ a11<\/sub>\u00a0+ a12<\/sub>\u00a0+ a13<\/sub>\u00a0+ a14<\/sub>. T\u00ednh b – c<\/p>\n <\/p>\n C\u00e2u 6:<\/b>\u00a0T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c<\/p>\n <\/p>\n A. 0,01\u00a0\u00a0\u00a0B. 0,1 \u00a0\u00a0\u00a0C. 1\u00a0\u00a0\u00a0D. 10<\/p>\n C\u00e2u 7:<\/b>\u00a0\u0110\u1eb7t a = log2<\/sub>3, b = log3<\/sub>5. H\u00e3y t\u00ednh bi\u1ec3u th\u1ee9c P = log6<\/sub>60 theo a v\u00e0 b<\/p>\n <\/p>\n C\u00e2u 8:<\/b><\/p>\n <\/p>\n C\u00e2u 9:<\/b>\u00a0Bi\u1ebft log(xy3<\/sup>) = 1 v\u00e0 log(x2<\/sup>y) = 1. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a log(xy)<\/p>\n A. 3\/5 \u00a0\u00a0\u00a0B. -1\/2\u00a0\u00a0\u00a0C. 1\/2\u00a0\u00a0\u00a0D. 1<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n 3 + 2log2<\/sub>x = log2<\/sub>y <=> log2<\/sub>23<\/sup>\u00a0+ log2<\/sub>x2<\/sup>\u00a0= log2<\/sub>y<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n B<\/p>\n C\u00e2u 2:<\/b><\/p>\n x = (log8<\/sub>2)log2<\/sub>8<\/sup>\u00a0= (log23<\/sup><\/sub>2)log2<\/sub>23<\/sup>\u00a0= (1\/3)3<\/sup>\u00a0= 3-3<\/sup>\u00a0=> log3<\/sub>x = -3<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A<\/p>\n C\u00e2u 3:<\/b><\/p>\n pH = -log[H+<\/sup>]<\/p>\n => [H+<\/sup>] = 10-pH<\/sup>\u00a0= 10-2,44<\/sup>\u00a0\u2248 0,00363 \u2248 3,6.10-3<\/sup>\u00a0(mol\/L).<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n C<\/p>\n C\u00e2u 4:<\/b><\/p>\n P = loga – logb + logb – logc + logc – logd + logd – loga – logy + logd + logx = log(x\/y)<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n B.<\/p>\n C\u00e2u 5:<\/b><\/p>\n <\/p>\n b – c = (log2002<\/sub>2 + log2002<\/sub>3 + log20002<\/sub>4 + log2002<\/sub>5) – (log2002<\/sub>10 + log2002<\/sub>11 + log2002<\/sub>12 + log2002<\/sub>13 + log2002<\/sub>14 )<\/p>\n <\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n B<\/p>\n C\u00e2u 6:<\/b><\/p>\n Bi\u1ec3u th\u1ee9c \u0111\u00e3 cho b\u1eb1ng<\/p>\n log100!<\/sub>2 + log100!<\/sub>3 + log100!<\/sub>4 + … + log100!<\/sub>100 = log100!<\/sub>(2.3.4….10) = log100!<\/sub>100! = 1<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n C<\/p>\n C\u00e2u 7:<\/b><\/p>\n <\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n D<\/p>\n C\u00e2u 8:<\/b><\/p>\n <\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n C\u00e2u 9:<\/b><\/p>\n Ta c\u00f3: 1 = log(xy3<\/sup>) = logx + 3logy v\u00e0 1 = loh(x2<\/sup>y) = 2logx + logy<\/p>\n T\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u01b0\u1ee3c logx = 2\/5 v\u00e0 logy = 1\/5 . V\u1eady log(xy) = logx + logy = 3\/5 .<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A<\/p>\n C\u00e1ch kh\u00e1c: T\u1eeb hai \u0111i\u1ec1u ki\u1ec7n \u0111\u00e3 cho, c\u00f3 xy3<\/sup>\u00a0= 10 v\u00e0 x2<\/sup>y = 10. T\u1eeb \u0111\u00f3, t\u00ednh \u0111\u01b0\u1ee3c<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Bi\u1ebft 3 + 2log2x = log2y . H\u00e3y bi\u1ec3u th\u1ecb y theo x A. y = 2x+3 \u00a0\u00a0\u00a0B. y = 8×2 \u00a0\u00a0\u00a0C. y = x2+8 \u00a0\u00a0\u00a0D. y = 3×2 C\u00e2u 2:\u00a0N\u1ebfu x = (log82)log28\u00a0th\u00ec log3x b\u1eb1ng: A. -3 \u00a0\u00a0\u00a0B. -1\/3\u00a0\u00a0\u00a0C. 1\/3 \u00a0\u00a0\u00a0D. 3 C\u00e2u 3:\u00a0\u0110\u1ed9 pH c\u1ee7a m\u1ed9t ch\u1ea5t \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi c\u00f4ng […]<\/p>\n","protected":false},"author":3,"featured_media":27531,"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 1-B<\/td>\n 2-A<\/td>\n 3-C<\/td>\n 4-B<\/td>\n 5-B<\/td>\n 6-C<\/td>\n 7-D<\/td>\n 8-A<\/td>\n 9-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n