C\u00e2u 5:<\/b><\/p>\n
<\/p>\n
C\u00e2u 6:<\/b>\u00a0Trong c\u00e1c h\u00e0m s\u1ed1 sau h\u00e0m s\u1ed1 n\u00e0o kh\u00f4ng ph\u1ea3i l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x) = cosxsinx ?<\/p>\n
<\/p>\n
C\u00e2u 7:<\/b>\u00a0T\u00ecm I=\u222b(3x2<\/sup>\u00a0– x + 1)ex<\/sup>dx<\/p>\n A. I = (3x2<\/sup>\u00a0– 7x +8)ex<\/sup>\u00a0+ C \u00a0\u00a0\u00a0 B. I = (3x2<\/sup>\u00a0– 7x)ex<\/sup>\u00a0+ C<\/p>\n C. I = (3x2<\/sup>\u00a0– 7x +8) + ex<\/sup>\u00a0+ C \u00a0\u00a0\u00a0D. I = (3x2<\/sup>\u00a0– 7x + 3)ex<\/sup>\u00a0+ C<\/p>\n C\u00e2u 8:<\/b><\/p>\n <\/p>\n C\u00e2u 9:<\/b>\u00a0M\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng v\u1edbi v\u1eadn t\u1ed1c v(t) (m\/s) c\u00f3 gia t\u1ed1c<\/p>\n <\/p>\n V\u1eadn t\u1ed1c ban \u0111\u1ea7u c\u1ee7a v\u1eadt l\u00e0 6m\/s. V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt sau 10 gi\u00e2y x\u1ea5p x\u1ec9 b\u1eb1ng<\/p>\n A. 10m\/s\u00a0\u00a0\u00a0 B. 11m\/s\u00a0\u00a0\u00a0 C. 12m\/s\u00a0\u00a0\u00a0 D. 13m\/s.<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 5:<\/b><\/p>\n \u0110\u1eb7t u = ex<\/sup>\u00a0+ 1 => u’ = ex<\/sup>. Ta c\u00f3<\/p>\n <\/p>\n C\u00e2u 6:<\/b><\/p>\n C\u00e1ch 1.<\/p>\n <\/p>\n C\u00e1ch 2. S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p bi\u1ebfn \u0111\u1ed5i s\u1ed1 ta c\u00f3:<\/p>\n \u0110\u1eb7t u = cosx th\u00ec u\u2019 = -sinx v\u00e0 \u222bsinxcosxdx = -\u222bu.u’dx = -\u222budu<\/p>\n <\/p>\n V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D.<\/p>\n C\u00e2u 7:<\/b><\/p>\n S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00ednh nguy\u00ean h\u00e0m t\u1eebng ph\u1ea7n ta c\u00f3:<\/p>\n \u0110\u1eb7t u = 3x2<\/sup>\u00a0– x + 1 v\u00e0 dv = ex<\/sup>dx ta c\u00f3 du = (6x – 1)dx v\u00e0 v = ex<\/sup>\u00a0. Do \u0111\u00f3:<\/p>\n \u222b(3x2<\/sup>\u00a0– x + 1)ex<\/sup>dx = (3x2<\/sup>\u00a0– x + 1)ex<\/sup>\u00a0– \u222b(6x – 1)ex<\/sup>dx<\/p>\n \u0110\u1eb7t u1<\/sub>\u00a0= 6x – 1; dv1<\/sub>\u00a0= ex<\/sup>dx Ta c\u00f3: du1<\/sub>\u00a0= 6dx v\u00e0 v1<\/sub>\u00a0= ex<\/sup>\u00a0.<\/p>\n Do \u0111\u00f3 \u222b(6x – 1)ex<\/sup>dx = (6x – 1)ex<\/sup>\u00a0– 6\u222bex<\/sup>dx = (6x – 1)ex<\/sup>\u00a0– 6ex<\/sup>\u00a0+ C<\/p>\n T\u1eeb \u0111\u00f3 suy ra<\/p>\n \u222b(3x2<\/sup>\u00a0– x + 1)ex<\/sup>dx = (3x2<\/sup>\u00a0– x + 1)ex<\/sup>\u00a0– (6x – 7)ex<\/sup>\u00a0+ C = (3x2<\/sup>\u00a0– 7x + 8)ex<\/sup>\u00a0+ C<\/p>\n V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n C\u00e2u 8:<\/b><\/p>\n <\/p>\n <\/p>\n V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n C\u00e2u 9:<\/b><\/p>\n V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt b\u1eb1ng<\/p>\n <\/p>\n v\u1edbi t= 0 ta c\u00f3 C = v(0) = 6 khi \u0111\u00f3 v(10)\u2248 13.<\/p>\n V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 5: C\u00e2u 6:\u00a0Trong c\u00e1c h\u00e0m s\u1ed1 sau h\u00e0m s\u1ed1 n\u00e0o kh\u00f4ng ph\u1ea3i l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x) = cosxsinx ? C\u00e2u 7:\u00a0T\u00ecm I=\u222b(3×2\u00a0– x + 1)exdx A. I = (3×2\u00a0– 7x +8)ex\u00a0+ C \u00a0\u00a0\u00a0 B. I = (3×2\u00a0– 7x)ex\u00a0+ C C. I = (3×2\u00a0– 7x +8) + ex\u00a0+ C \u00a0\u00a0\u00a0D. I = (3×2\u00a0– […]<\/p>\n","protected":false},"author":3,"featured_media":27077,"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 5-D<\/td>\n 6-D<\/td>\n 7-A<\/td>\n 8-C<\/td>\n 9-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n