C\u00e2u 1:<\/b>\u00a0Gi\u1ea3 s\u1eed x l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n
<\/p>\n
A. 0\u00a0\u00a0\u00a0B. ln3 \u00a0\u00a0\u00a0C. \u2013ln3 \u00a0\u00a0\u00a0D. 1\/ln3<\/p>\n
C\u00e2u 2:<\/b>\u00a0T\u00ednh t\u00edch t\u1ea5t c\u1ea3 c\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 32×2\u00a0+ 2x + 1<\/sup>\u00a0– 28.3x2\u00a0+ x<\/sup>\u00a0+ 9 = 0<\/p>\n A. -4 \u00a0\u00a0\u00a0B. -2 \u00a0\u00a0\u00a0C. 2 \u00a0\u00a0\u00a0D. 4<\/p>\n C\u00e2u 3:<\/b>\u00a0T\u00ecm nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 2x – 1<\/sup>\u00a0= 31 – 2x<\/sup><\/p>\n <\/p>\n C\u00e2u 4:<\/b>\u00a0Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh (x2<\/sup>\u00a0– 2x)lnx = lnx3<\/sup><\/p>\n A. x = 1, x = 3 \u00a0\u00a0\u00a0B. x = -1, x = 3 \u00a0\u00a0\u00a0 C. x = \u00b11, x = 3 \u00a0\u00a0\u00a0D. x = 3<\/p>\n C\u00e2u 5:<\/b>\u00a0N\u1ebfu log7<\/sub>(log3<\/sub>(log2<\/sub>x)) = 0 th\u00ec x-1\/2<\/sup>\u00a0b\u1eb1ng :<\/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 \u0110\u1ec3 \u00fd r\u1eb1ng<\/p>\n <\/p>\n n\u00ean ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi<\/p>\n <\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n C\u00e2u 2:<\/b><\/p>\n \u0110\u1eb7t t = 3x2<\/sup>\u00a0+ x > 0 nh\u1eadn \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n <\/p>\n V\u1edbi t = 1\/3 = 3-1<\/sup>\u00a0\u0111\u01b0\u1ee3c 3x2\u00a0+ x<\/sup>\u00a0= 3-1<\/sup>\u00a0<=> x2<\/sup>\u00a0+ x + 1 = 0(v\u00f4 nghi\u1ec7m)<\/p>\n V\u1edbi t = 9 \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh x2<\/sup>\u00a0+ x – 2 = 0 <=> x -2 ho\u1eb7c x = 1<\/p>\n T\u00edch c\u1ee7a hai nghi\u1ec7m n\u00e0y b\u1eb1ng -2.<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n B<\/p>\n C\u00e2u 3:<\/b><\/p>\n C\u00f3 nhi\u1ec1u c\u00e1ch bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y. Tuy nhi\u00ean, nh\u1eadn th\u1ea5y c\u00e1c bi\u1ec3u th\u1ee9c trong c\u00e1c ph\u01b0\u01a1ng \u00e1n \u0111\u1ec1u ch\u1ee9a log2<\/sub>3 , n\u00ean ta l\u1ea5y l\u00f4garit c\u01a1 s\u1ed1 2 hai v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 nh\u1eadn \u0111\u01b0\u1ee3c (x – 1) = (1 – 2x)log2<\/sub>3<\/p>\n <=> x(2log2<\/sub>3 + 1) = log2<\/sub>3 + 1<\/p>\n <\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n D<\/p>\n C\u00e2u 4:<\/b><\/p>\n \u0110i\u1ec1u ki\u1ec7n x > 0. Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi<\/p>\n (x2<\/sup>\u00a0-2x)lnx = 3lnx <=> (x2<\/sup>\u00a0– 2x + 3)lnx = 0<\/p>\n <\/p>\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 x = 1, x = 3 .<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n Ch\u00fa \u00fd. Sai l\u1ea7m th\u01b0\u1eddng g\u1eb7p l\u00e0 qu\u00ean \u0111i\u1ec1u ki\u1ec7n d\u1eabn \u0111\u1ebfn kh\u00f4ng lo\u1ea1i \u0111\u01b0\u1ee3c nghi\u1ec7m x = -1 v\u00e0 ch\u1ecdn ph\u01b0\u01a1ng \u00e1n nhi\u1ec5u C.<\/p>\n Th\u1eadm ch\u00ed, c\u00f3 th\u1ec3 h\u1ecdc sinh bi\u1ebfn \u0111\u1ed5i (x2<\/sup>\u00a0– 2x)lnx = 3lnx <=> x2<\/sup>\u00a0-2x = 3(gi\u1ea3n \u01b0\u1edbc cho lnx) d\u1eabn \u0111\u1ebfn m\u1ea5t nghi\u1ec7m x = 1 v\u00e0 ch\u1ecdn ph\u01b0\u01a1ng \u00e1n nhi\u1ec5u D.<\/p>\n C\u00e2u 5:<\/b><\/p>\n log7<\/sub>(log3<\/sub>(log2<\/sub>x)) = 0 <=> log3<\/sub>(log2<\/sub>x) = 70<\/sup>\u00a0= 1<\/p>\n <=> log2<\/sub>x = 3t<\/sup>\u00a0<=> x = 23<\/sup>\u00a0= 8<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n C<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Gi\u1ea3 s\u1eed x l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh A. 0\u00a0\u00a0\u00a0B. ln3 \u00a0\u00a0\u00a0C. \u2013ln3 \u00a0\u00a0\u00a0D. 1\/ln3 C\u00e2u 2:\u00a0T\u00ednh t\u00edch t\u1ea5t c\u1ea3 c\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 32×2\u00a0+ 2x + 1\u00a0– 28.3×2\u00a0+ x\u00a0+ 9 = 0 A. -4 \u00a0\u00a0\u00a0B. -2 \u00a0\u00a0\u00a0C. 2 \u00a0\u00a0\u00a0D. 4 C\u00e2u 3:\u00a0T\u00ecm nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh 2x – 1\u00a0= 31 – 2x […]<\/p>\n","protected":false},"author":3,"featured_media":27406,"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-A<\/td>\n 2-B<\/td>\n 3-D<\/td>\n 4-D<\/td>\n 5-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n