C\u00e2u 1:<\/b>\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh 10x<\/sup>\u00a0= 0,00001<\/p>\n A. x = -log4 \u00a0\u00a0\u00a0B. x = -log5 \u00a0\u00a0\u00a0C. x = -4 \u00a0\u00a0\u00a0D. x = -5<\/p>\n C\u00e2u 2:<\/b>\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n <\/p>\n C\u00e2u 3:<\/b>\u00a0Cho ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n <\/p>\n Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y n\u1eb1m trong kho\u1ea3ng n\u00e0o d\u01b0\u1edbi \u0111\u00e2y ?<\/p>\n <\/p>\n C\u00e2u 4:<\/b>\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh 32x – 3<\/sup>\u00a0= 7 . Vi\u1ebft nghi\u1ec7m d\u01b0\u1edbi d\u1ea1ng th\u1eadp ph\u00e2n, l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n ngh\u00ecn.<\/p>\n A. x = 2,38\u00a0\u00a0\u00a0B. x = 2,386 \u00a0\u00a0\u00a0C. x = 2,384\u00a0\u00a0\u00a0D. x = 1,782<\/p>\n C\u00e2u 5:<\/b>\u00a0T\u00ednh t\u1ed5ng b\u00ecnh ph\u01b0\u01a1ng c\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 4x2\u00a0+ 2<\/sup>\u00a0– 9.2x2\u00a0+ 2<\/sup>\u00a0+ 8 = 0<\/p>\n A. 2\u00a0\u00a0\u00a0B. 4\u00a0\u00a0\u00a0C. 17\u00a0\u00a0\u00a0D. 65<\/p>\n C\u00e2u 6:<\/b>\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh 4x<\/sup>\u00a0+ 2x + 1<\/sup>\u00a0– 15 = 0. Vi\u1ebft nghi\u1ec7m t\u00ecm \u0111\u01b0\u1ee3c d\u01b0\u1edbi d\u1ea1ng th\u1eadp ph\u00e2n, l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m<\/p>\n A. x = 0,43 \u00a0\u00a0\u00a0 B. x = 0,63 \u00a0\u00a0\u00a0C. x = 1,58 \u00a0\u00a0\u00a0D. x = 2,32<\/p>\n C\u00e2u 7:<\/b>\u00a0Gi\u1ea3 s\u1eed x1<\/sub>, x2<\/sub>\u00a0l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 7x<\/sup>\u00a0+ 2.71 – x<\/sup>\u00a0– 9 = 0.<\/p>\n <\/p>\n A. log2<\/sub>7 + 1\u00a0\u00a0\u00a0B. log7<\/sub>2 + 1 \u00a0\u00a0\u00a0C. log7<\/sub>2 \u00a0\u00a0\u00a0D. log2<\/sub>7<\/p>\n C\u00e2u 8:<\/b>\u00a0T\u00ecm nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 41 – x<\/sup>\u00a0= 32x + 1<\/sup><\/p>\n <\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n 10x<\/sup>\u00a0= 0,00001 <=> 10x<\/sup>\u00a0= 10-5<\/sup>\u00a0<=> x = -5<\/p>\n C\u00e2u 2:<\/b><\/p>\n PT <=> 1 – e-2x<\/sup>\u00a0= 1\/2 <=> e-2x<\/sup>\u00a0= 1\/2 <=> -2x = ln(1\/2) = -ln2 <=> x = (1\/2)ln2<\/p>\n C\u00e2u 3:<\/b><\/p>\n PT <=> 5x – 1<\/sup>\u00a0= 5-2x<\/sup>\u00a0<=> x – 1 = -2x <=> 3x = 1 <=> x = 1\/3<\/p>\n C\u00e2u 4:<\/b><\/p>\n 32x – 3<\/sup>\u00a0= 7 <=> 2x – 3 = log3<\/sub>7 <=> x = (1\/2)(log3<\/sub>7 + 3) \u2248 2,386<\/p>\n C\u00e2u 5:<\/b><\/p>\n \u0110\u1eb7t t = 2x2\u00a0+ 2<\/sup>, nh\u1eadn \u0111\u01b0\u1ee3c t2<\/sup>\u00a0– 9t + 8 = 0 <=> t = 1 ho\u1eb7c t = 8<\/p>\n V\u1edbi t = 1, nh\u1eadn \u0111\u01b0\u1ee3c 2x2\u00a0+ 2<\/sup>\u00a0= 1 (v\u00f4 nghi\u1ec7m)<\/p>\n V\u1edbi t = 8, ta c\u00f3: 2x2\u00a0+ 2<\/sup>\u00a0= 8 = 23<\/sup>\u00a0<=> x2<\/sup>\u00a0+ 2 = 3 <=> x = \u00b1 1<\/p>\n T\u1ed5ng b\u00ecnh ph\u01b0\u01a1ng c\u00e1c nghi\u1ec7m: 12<\/sup>\u00a0+ (-1)2<\/sup>\u00a0= 2<\/p>\n C\u00e2u 6:<\/b><\/p>\n \u0110\u1eb7t t = 2x<\/sup>\u00a0> 0, nh\u1eadn \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh t2<\/sup>\u00a0+ 2t – 15 = 0 <=> t = -5 (lo\u1ea1i) ho\u1eb7c t = 3 hay 2x<\/sup>\u00a0= 3 <=> x = log2<\/sub>3 = ln3\/ln2 \u2248 1,58<\/p>\n C\u00e2u 7:<\/b><\/p>\n \u0110\u1eb7t t = 7x<\/sup>, \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n <\/p>\n T\u1eeb \u0111\u00f3 t\u00ecm \u0111\u01b0\u1ee3c x1<\/sub>\u00a0= 1, x2<\/sub>\u00a0= log7<\/sub>2. Ta c\u00f3:<\/p>\n <\/p>\n C\u00e2u 8:<\/b><\/p>\n 41 – x<\/sup>\u00a0= 32x + 1<\/sup>\u00a0<=> 22 – 2x<\/sup>\u00a0= 32x + 1<\/sup><\/p>\n L\u1ea5y l\u00f4garit c\u01a1 s\u1ed1 3 hai v\u1ebf ta \u0111\u01b0\u1ee3c :<\/p>\n (2 – 2x)log3<\/sub>2 = 2x + 1 <=> 2x(log3<\/sub>2 + 1) = 2log3<\/sub>2 – 1<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh 10x\u00a0= 0,00001 A. x = -log4 \u00a0\u00a0\u00a0B. x = -log5 \u00a0\u00a0\u00a0C. x = -4 \u00a0\u00a0\u00a0D. x = -5 C\u00e2u 2:\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh C\u00e2u 3:\u00a0Cho ph\u01b0\u01a1ng tr\u00ecnh Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y n\u1eb1m trong kho\u1ea3ng n\u00e0o d\u01b0\u1edbi \u0111\u00e2y ? C\u00e2u 4:\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh 32x – 3\u00a0= 7 . Vi\u1ebft nghi\u1ec7m d\u01b0\u1edbi d\u1ea1ng […]<\/p>\n","protected":false},"author":3,"featured_media":27390,"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-D<\/td>\n 2-B<\/td>\n 3-A<\/td>\n 4-B<\/td>\n 5-A<\/td>\n 6-C<\/td>\n 7-D<\/td>\n 8-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n