C\u00e2u 1:<\/b>\u00a0Gi\u1ea3 s\u1eed<\/p>\n
<\/p>\n
Gi\u00e1 tr\u1ecb c\u1ee7a K l\u00e0:<\/p>\n
A.9 \u00a0\u00a0\u00a0 B.3\u00a0\u00a0\u00a0 C.81\u00a0\u00a0\u00a0 D.8<\/p>\n
C\u00e2u 2:<\/b>\u00a0Cho:<\/p>\n
<\/p>\n
T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a a-b.<\/p>\n
A.3 \u00a0\u00a0\u00a0 B.1 \u00a0\u00a0\u00a0C.2 \u00a0\u00a0\u00a0D.0.<\/p>\n
C\u00e2u 3:<\/b>\u00a0Cho<\/p>\n
<\/p>\n
Gi\u1ea3 s\u1eed \u0111\u1eb7t t = \u221bex<\/sup>\u00a0+ 2 th\u00ec ta \u0111\u01b0\u1ee3c:<\/p>\n <\/p>\n C\u00e2u 4:<\/b>\u00a0Cho<\/p>\n <\/p>\n Khi \u0111\u00f3 a+b b\u1eb1ng<\/p>\n A.10+ \u221a7 \u00a0\u00a0\u00a0B.22\u00a0\u00a0\u00a0C. \u221a7 + 6 \u00a0\u00a0\u00a0 D.6.<\/p>\n C\u00e2u 5:<\/b>\u00a0Cho<\/p>\n <\/p>\n \u0110\u1eb7t t = x2<\/sup>\u00a0. Bi\u1ebft<\/p>\n <\/p>\n C\u00e2u 6:<\/b>\u00a0N\u1ebfu<\/p>\n <\/p>\n v\u1edbi a < d < b th\u00ec<\/p>\n <\/p>\n A.-2 \u00a0\u00a0\u00a0B.3 \u00a0\u00a0\u00a0 C.8 \u00a0\u00a0\u00a0 D.0<\/p>\n C\u00e2u 7:<\/b>\u00a0Cho t\u00edch ph\u00e2n<\/p>\n <\/p>\n N\u1ebfu bi\u1ebfn \u0111\u1ed5i s\u1ed1 t = sin2<\/sup>x th\u00ec:<\/p>\n <\/p>\n C\u00e2u 8:<\/b>\u00a0Bi\u1ebft<\/p>\n <\/p>\n Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y nh\u1eadn gi\u00e1 tr\u1ecb \u0111\u00fang?<\/p>\n <\/p>\n C\u00e2u 9:<\/b><\/p>\n <\/p>\n A. Kh\u00f4ng x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c \u00a0\u00a0\u00a0 B.1<\/p>\n C.3 \u00a0\u00a0\u00a0 D.-1<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 2:<\/b><\/p>\n <\/p>\n Khi x = 1 th\u00ec t = e, khi x = e th\u00ec t = ee<\/sup>\u00a0+ 1 .<\/p>\n <\/p>\n T\u1eeb \u0111\u00f3 suy ra a \u2013 b = 0.<\/p>\n C\u00e2u 3:<\/b><\/p>\n <\/p>\n =>(t – 2)3<\/sup>\u00a0= ex<\/sup>\u00a0=> 3(t-2)2<\/sup>dt = ex<\/sup>dx<\/p>\n Khi \u0111\u00f3<\/p>\n <\/p>\n C\u00e2u 4:<\/b><\/p>\n Ta c\u00f3<\/p>\n <\/p>\n C\u00e2u 5:<\/b><\/p>\n \u0110\u1eb7t t = x2<\/sup>\u00a0=7gt; dt = 2xdx. Ta c\u00f3:<\/p>\n <\/p>\n C\u00e2u 6:<\/b><\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n C\u00e2u 7:<\/b><\/p>\n \u0110\u1eb7t t = sin2<\/sup>x th\u00ec khi \u0111\u00f3<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Gi\u1ea3 s\u1eed Gi\u00e1 tr\u1ecb c\u1ee7a K l\u00e0: A.9 \u00a0\u00a0\u00a0 B.3\u00a0\u00a0\u00a0 C.81\u00a0\u00a0\u00a0 D.8 C\u00e2u 2:\u00a0Cho: T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a a-b. A.3 \u00a0\u00a0\u00a0 B.1 \u00a0\u00a0\u00a0C.2 \u00a0\u00a0\u00a0D.0. C\u00e2u 3:\u00a0Cho Gi\u1ea3 s\u1eed \u0111\u1eb7t t = \u221bex\u00a0+ 2 th\u00ec ta \u0111\u01b0\u1ee3c: C\u00e2u 4:\u00a0Cho Khi \u0111\u00f3 a+b b\u1eb1ng A.10+ \u221a7 \u00a0\u00a0\u00a0B.22\u00a0\u00a0\u00a0C. \u221a7 + 6 \u00a0\u00a0\u00a0 D.6. C\u00e2u 5:\u00a0Cho \u0110\u1eb7t t […]<\/p>\n","protected":false},"author":3,"featured_media":26976,"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-A<\/td>\n 4-B<\/td>\n 5-B<\/td>\n 6-B<\/td>\n 7-A<\/td>\n 8-D<\/td>\n 9-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n