C\u00e2u 1:<\/b>\u00a0\u0110a\u0323o ha\u0300m cu\u0309a ha\u0300m s\u00f4\u0301 y = (x2<\/sup>\u00a0+ 1)e3x<\/sup>\u00a0la\u0300<\/p>\n A. 2x.e3x<\/sup>\u00a0\u00a0\u00a0B. e3x<\/sup>(3x2<\/sup>\u00a0+ 2x + 3)\u00a0\u00a0\u00a0C. 3(x2<\/sup>\u00a0+ 1)e3x<\/sup>\u00a0\u00a0\u00a0D. 6xe3x<\/sup><\/p>\n C\u00e2u 2:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301<\/p>\n <\/p>\n A. 0 \u00a0\u00a0\u00a0B. \u20133\u00a0\u00a0\u00a0C. 6 \u00a0\u00a0\u00a0D. -6<\/p>\n C\u00e2u 3:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 6x2<\/sup>\u00a0+ 9x + 7. Ha\u0300m s\u00f4\u0301 \u0111\u00f4\u0300ng bi\u00ea\u0301n tr\u00ean khoa\u0309ng<\/p>\n A. (1; 3)\u00a0\u00a0\u00a0B. (1; +\u221e)\u00a0\u00a0\u00a0C. (-\u221e; 3) \u00a0\u00a0\u00a0D. (-\u221e; 1) va\u0300 (3; +\u221e)<\/p>\n C\u00e2u 4:<\/b>\u00a0Ha\u0300m s\u00f4\u0301<\/p>\n <\/p>\n nghi\u0323ch bi\u00ea\u0301n tr\u00ean ca\u0301c khoa\u0309ng<\/p>\n A. (0; 2) va\u0300 (2; +\u221e) \u00a0\u00a0\u00a0B. (-\u221e; 0) va\u0300 (2; +\u221e)<\/p>\n C. [0; 1) va\u0300 (1; 2] \u00a0\u00a0\u00a0D. (0; 1) va\u0300 (1; 2)<\/p>\n C\u00e2u 5:<\/b>\u00a0Ha\u0300m s\u00f4\u0301 na\u0300o trong ca\u0301c ha\u0300m s\u00f4\u0301 sau \u0111\u00e2y \u0111\u00f4\u0300ng bi\u00ea\u0301n tr\u00ean R<\/p>\n A. y = x2<\/sup>\u00a0– 2x + 3\u00a0\u00a0\u00a0B. y = x3<\/sup>\u00a0+ x<\/p>\n C. y = 1 + 1\/(x2<\/sup>\u00a0+ 1)\u00a0\u00a0\u00a0 D. y = ln(x2<\/sup>\u00a0+ 1)<\/p>\n C\u00e2u 6:<\/b>\u00a0Ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ mx + m \u0111\u00f4\u0300ng bi\u00ea\u0301n tr\u00ean (-\u221e; +\u221e) khi va\u0300 chi\u0309 khi<\/p>\n A. m = 3\u00a0\u00a0\u00a0B. m \u2265 3\u00a0\u00a0\u00a0C. m \u2264 3\u00a0\u00a0\u00a0D. 0 \u2264 m \u2264 3<\/p>\n C\u00e2u 7:<\/b>\u00a0Ha\u0300m s\u00f4\u0301 y = 2x3<\/sup>\u00a0– 9x2<\/sup>\u00a0+ 12x – 4<\/p>\n A. Co\u0301 c\u01b0\u0323c \u0111a\u0323i ma\u0300 kh\u00f4ng co\u0301 c\u01b0\u0323c ti\u00ea\u0309u\u00a0\u00a0\u00a0B. Co\u0301 c\u01b0\u0323c ti\u00ea\u0309u ma\u0300 kh\u00f4ng co\u0301 c\u01b0\u0323c \u0111a\u0323i<\/p>\n C. Kh\u00f4ng co\u0301 c\u01b0\u0323c \u0111a\u0323i va\u0300 c\u01b0\u0323c ti\u00ea\u0309u\u00a0\u00a0\u00a0D. Co\u0301 ca\u0309 c\u01b0\u0323c \u0111a\u0323i va\u0300 c\u01b0\u0323c ti\u00ea\u0309u.<\/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: y’ = (x2<\/sup>\u00a0+ 1)’.33x<\/sup>\u00a0+ (x2<\/sup>\u00a0+ 1)(e3x<\/sup>)’ = 2x.e3x<\/sup>\u00a0+ 3(x2<\/sup>\u00a0+ 1)e3x<\/sup>\u00a0= (3x2<\/sup>\u00a0+ 2x + 3)e3x<\/sup><\/p>\n C\u00e2u 2:<\/b><\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n C\u00e2u 3:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 12x + 9 = 3(x – 1)(x – 3), y’ > 0 <=> x \u2208(-\u221e; 1) \u222a (3; +\u221e)<\/p>\n V\u1eady h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; 1) v\u00e0 (3; +\u221e) .<\/p>\n C\u00e2u 4:<\/b><\/p>\n T\u1eadp x\u00e1c \u0111\u1ecbnh: D = R\\{1}. Ta c\u00f3<\/p>\n <\/p>\n y’ < 0 <=> x \u2208 (0; 1) \u222a (1; 2)<\/p>\n V\u1eady h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng [0; 1 ) v\u00e0 (1; 2]<\/p>\n C\u00e2u 5:<\/b><\/p>\n H\u00e0m s\u1ed1 y = x2<\/sup>\u00a0– 2x + 3 c\u00f3 \u0111\u1ed3 th\u1ecb l\u00e0 parabol n\u00ean kh\u00f4ng th\u1ec3 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R<\/p>\n H\u00e0m s\u1ed1 y = x3<\/sup>\u00a0+ x x\u00e1c \u0111\u1ecbnh tr\u00ean R v\u00e0 c\u00f3 \u0111\u1ea1o h\u00e0m y’ = 3x2<\/sup>\u00a0+ 1 > 0, \u2200x \u2208 R n\u00ean \u0111\u1ed3ng bi\u1ebfn tr\u00ean R .<\/p>\n H\u00e0m s\u1ed1 y = x2<\/sup>\u00a0+ 1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean (0; +\u221e) n\u00ean h\u00e0m s\u1ed1<\/p>\n <\/p>\n ngh\u1ecbch bi\u1ebfn tr\u00ean (0; +\u221e). Do \u0111\u00f3 h\u00e0m s\u1ed1<\/p>\n <\/p>\n kh\u00f4ng th\u1ec3 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R .<\/p>\n H\u00e0m s\u1ed1 y = x2<\/sup>\u00a0+ 1 ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; 0) n\u00ean h\u00e0m s\u1ed1 y = ln(x2<\/sup>\u00a0+ 1) ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; 0) . Do \u0111\u00f3 h\u00e0m s\u1ed1 y = ln(x2<\/sup>\u00a0+ 1) kh\u00f4ng th\u1ec3 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R .<\/p>\n C\u00e2u 6:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 6x + m. H\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ mx + m \u0111\u1ed3ng bi\u1ebfn tr\u00ean (-\u221e; +\u221e) khi v\u00e0 ch\u1ec9 khi y’ \u2265 0, \u2200x \u2208 (-\u221e; +\u221e) <=> m \u2265 3<\/p>\n C\u00e2u 7:<\/b><\/p>\n Ta c\u00f3: y’ = 6x2<\/sup>\u00a0– 18x + 12 = 6(x – 1)(x – 2). L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean ta suy ra h\u00e0m s\u1ed1 c\u00f3 c\u1ea3 c\u1ef1c \u0111\u1ea1i v\u00e0 c\u1ef1c ti\u1ec3u.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0\u0110a\u0323o ha\u0300m cu\u0309a ha\u0300m s\u00f4\u0301 y = (x2\u00a0+ 1)e3x\u00a0la\u0300 A. 2x.e3x\u00a0\u00a0\u00a0B. e3x(3×2\u00a0+ 2x + 3)\u00a0\u00a0\u00a0C. 3(x2\u00a0+ 1)e3x\u00a0\u00a0\u00a0D. 6xe3x C\u00e2u 2:\u00a0Cho ha\u0300m s\u00f4\u0301 A. 0 \u00a0\u00a0\u00a0B. \u20133\u00a0\u00a0\u00a0C. 6 \u00a0\u00a0\u00a0D. -6 C\u00e2u 3:\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3\u00a0– 6×2\u00a0+ 9x + 7. Ha\u0300m s\u00f4\u0301 \u0111\u00f4\u0300ng bi\u00ea\u0301n tr\u00ean khoa\u0309ng A. (1; 3)\u00a0\u00a0\u00a0B. (1; +\u221e)\u00a0\u00a0\u00a0C. (-\u221e; 3) \u00a0\u00a0\u00a0D. […]<\/p>\n","protected":false},"author":3,"featured_media":26603,"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-D<\/td>\n 3-D<\/td>\n 4-C<\/td>\n 5-B<\/td>\n 6-B<\/td>\n 7-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n