C\u00e2u 1:<\/b>\u00a0Vi\u1ebft c\u00e1c s\u1ed1<\/p>\n
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theo th\u1ee9 t\u1ef1 t\u0103ng d\u1ea7n<\/p>\n
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C\u00e2u 2:<\/b>\u00a0T\u00ecm \u0111\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 y = log5<\/sub>(xex<\/sup>)<\/p>\n <\/p>\n C\u00e2u 3:<\/b>\u00a0T\u00ecm c\u00e1c kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = x2<\/sup>e-4x<\/sup><\/p>\n <\/p>\n C\u00e2u 4:<\/b>\u00a0T\u00ecm c\u00e1c kho\u1ea3ng ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = 3ln(x +1) + x – x2<\/sup>\/2<\/p>\n A.(-1; 2)\u00a0\u00a0\u00a0C. (-2 ;-1) v\u00e0 (2; +\u221e)<\/p>\n B. (2; +\u221e)\u00a0\u00a0\u00a0D. (-\u221e; -2) v\u00e0 (-1 ;2)<\/p>\n C\u00e2u 5:<\/b>\u00a0Cho hai s\u1ed1 th\u1ef1c a v\u00e0 b , v\u1edbi 0 < a < b < 1. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang ?<\/p>\n A. logb<\/sub>a < 1 < loga<\/sub>b\u00a0\u00a0\u00a0C. loga<\/sub>b < 1 < logb<\/sub>a<\/p>\n B. logb<\/sub>a < loga<\/sub>b < 1 \u00a0\u00a0\u00a0D. 1 < loga<\/sub>b < logb<\/sub>a<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n Ta c\u00f3 -1 < 0 < \u221a2 < \u03c0 v\u00e0 0 < 1\/3 < 1 n\u00ean<\/p>\n <\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n C\u00e2u 2:<\/b><\/p>\n \u0110\u1ec3 thu\u1eadn ti\u1ec7n, ta vi\u1ebft l\u1ea1i<\/p>\n <\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n D<\/p>\n C\u00e2u 3:<\/b><\/p>\n T\u1eadp x\u00e1c \u0111\u1ecbnh R.<\/p>\n Ta c\u00f3:<\/p>\n y’ = 2xe-4x<\/sup>\u00a0+ x2<\/sup>e-4x<\/sup>(-4) = 2e-4x<\/sup>x(1 – 2x)<\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n Kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 (0; 1\/2) .<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n C<\/p>\n C\u00e2u 4:<\/b><\/p>\n T\u1eadp x\u00e1c \u0111\u1ecbnh : (-1; +\u221e)<\/p>\n <\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean :<\/p>\n <\/p>\n T\u1eeb \u0111\u00f3, kho\u1ea3ng ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 l\u00e0(2; +\u221e) .<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n B<\/p>\n C\u00e2u 5:<\/b><\/p>\n \u0110\u1eb7t c = b – a ta c\u00f3 c > 0. Do c\u00e1c h\u00e0m s\u1ed1 y = loga<\/sub>x v\u00e0 logb<\/sub>x ngh\u1ecbch bi\u1ebfn tr\u00ean (0; +\u221e) n\u00ean ta c\u00f3 loga<\/sub>b = loga<\/sub>(a + c) < loga<\/sub>a = 1 v\u00e0 logb<\/sub>a = logb<\/sub>(b – c) > logb<\/sub>b = 1.<\/p>\n V\u1eady loga<\/sub>b < a < logb<\/sub>a<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Vi\u1ebft c\u00e1c s\u1ed1 theo th\u1ee9 t\u1ef1 t\u0103ng d\u1ea7n C\u00e2u 2:\u00a0T\u00ecm \u0111\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 y = log5(xex) C\u00e2u 3:\u00a0T\u00ecm c\u00e1c kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = x2e-4x C\u00e2u 4:\u00a0T\u00ecm c\u00e1c kho\u1ea3ng ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = 3ln(x +1) + x – x2\/2 A.(-1; 2)\u00a0\u00a0\u00a0C. (-2 ;-1) v\u00e0 (2; +\u221e) […]<\/p>\n","protected":false},"author":3,"featured_media":27493,"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-D<\/td>\n 3-C<\/td>\n 4-B<\/td>\n 5-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n