C\u00e2u 1:<\/b>\u00a0T\u00ecm GTNN c\u1ee7a h\u00e0m s\u1ed1 y = x2<\/sup>\u00a0– 3x + 5<\/p>\n A. 3\/2 \u00a0\u00a0\u00a0B. 11\/4\u00a0\u00a0\u00a0 C. 3 \u00a0\u00a0\u00a0 D. 5<\/p>\n C\u00e2u 2:<\/b>\u00a0GTLN c\u1ee7a h\u00e0m s\u1ed1 y = sin2<\/sup>x – \u221a3cosx tr\u00ean \u0111o\u1ea1n [0; \u03c0] l\u00e0<\/p>\n A. 1\u00a0\u00a0\u00a0B. 7.4\u00a0\u00a0\u00a0 C. 2 \u00a0\u00a0\u00a0 D. 1\/4<\/p>\n C\u00e2u 3:<\/b>\u00a0GTNN c\u1ee7a h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 9x + 1 tr\u00ean \u0111o\u1ea1n [-4;4] l\u00e0<\/p>\n A. 0 \u00a0\u00a0\u00a0 B. 1 \u00a0\u00a0\u00a0 C. 4 \u00a0\u00a0\u00a0 D. -1<\/p>\n C\u00e2u 4:<\/b>\u00a0GTLN c\u1ee7a h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 8x2<\/sup>\u00a0+ 16 tr\u00ean \u0111o\u1ea1n [-1;3] l\u00e0<\/p>\n A. 0 \u00a0\u00a0\u00a0 B. 15 \u00a0\u00a0\u00a0 C. 25 \u00a0\u00a0\u00a0 D. 30<\/p>\n C\u00e2u 5:<\/b>\u00a0GTNN c\u1ee7a h\u00e0m s\u1ed1 y = x\/(x+2) tr\u00ean n\u1eeda kho\u1ea3ng (-2;4] l\u00e0<\/p>\n A. 0 \u00a0\u00a0\u00a0 B. 1 \u00a0\u00a0\u00a0 C.2\/3 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n C\u00e2u 6:<\/b>\u00a0GTNN c\u1ee7a h\u00e0m s\u1ed1 y = x + 2 + 1\/(x – 1) tr\u00ean kho\u1ea3ng (1; +\u221e) l\u00e0:<\/p>\n A. Kh\u00f4ng t\u1ed3n t\u1ea1i\u00a0\u00a0\u00a0B. 3\/2 \u00a0\u00a0\u00a0C. 2\u00a0\u00a0\u00a0D = 7\/4<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 2:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1 y = sin2<\/sup>x – \u221a3cosx tr\u00ean \u0111o\u1ea1n [0; \u03c0]<\/p>\n y’ = 2sinxcosx + \u221a3sinx = sinx(2cosx + \u221a3) .<\/p>\n =>y\u2019=0 <=> x=0 ho\u1eb7c x = \u03c0 ho\u1eb7c x = 5\u03c0\/6 .<\/p>\n y(0) = -\u221a3; y(\u03c0) = \u221a3; y(5\u03c0\/6) = 7\/4<\/p>\n C\u00e2u 3:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 9x + 1 tr\u00ean \u0111o\u1ea1n [-4;4].<\/p>\n Ta c\u00f3 y’ = 3x2<\/sup>\u00a0+ 6x – 9 => y’ = 0<\/p>\n <\/p>\n y(1) = -4, y(-3) = 28; y(4) = 77; y(-4) = 21<\/p>\n C\u00e2u 4:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 8x2<\/sup>\u00a0+ 16 tr\u00ean \u0111o\u1ea1n [-1;3]<\/p>\n y’ = 4x3<\/sup>\u00a0– 6x<\/p>\n <\/p>\n y(0) = 16, y(2) = 0; y(-1) = 9; y(3) = 25<\/p>\n C\u00e2u 5:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1<\/p>\n <\/p>\n Ta c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 GTNN<\/p>\n C\u00e2u 6:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1<\/p>\n <\/p>\n y’ = 0 => x = 2. B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 y=5.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0T\u00ecm GTNN c\u1ee7a h\u00e0m s\u1ed1 y = x2\u00a0– 3x + 5 A. 3\/2 \u00a0\u00a0\u00a0B. 11\/4\u00a0\u00a0\u00a0 C. 3 \u00a0\u00a0\u00a0 D. 5 C\u00e2u 2:\u00a0GTLN c\u1ee7a h\u00e0m s\u1ed1 y = sin2x – \u221a3cosx tr\u00ean \u0111o\u1ea1n [0; \u03c0] l\u00e0 A. 1\u00a0\u00a0\u00a0B. 7.4\u00a0\u00a0\u00a0 C. 2 \u00a0\u00a0\u00a0 D. 1\/4 C\u00e2u 3:\u00a0GTNN c\u1ee7a h\u00e0m s\u1ed1 y = x3\u00a0+ 3×2\u00a0– 9x + […]<\/p>\n","protected":false},"author":3,"featured_media":27986,"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-B<\/td>\n 3-A<\/td>\n 4-C<\/td>\n 5-D<\/td>\n 6-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n