C\u00e2u 8:<\/b>\u00a0Gi\u00e1 tr\u1ecb c\u1ee7a m\u1ed9t chi\u1ebfc xe \u00f4 t\u00f4 sau t n\u0103m k\u1ec3 t\u1eeb khi mua \u0111\u01b0\u1ee3c \u01b0\u1edbc l\u01b0\u1ee3ng b\u1eb1ng c\u00f4ng th\u1ee9c G(t) = 600e-0,12t<\/sup>\u00a0(tri\u1ec7u \u0111\u1ed3ng). T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a chi\u1ebfc xe n\u00e0y t\u1ea1i hai th\u1eddi \u0111i\u1ec3m : l\u00fac mua v\u00e0 l\u00fac \u0111\u00e3 s\u1eed d\u1ee5ng 5 n\u0103m (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng tri\u1ec7u)<\/p>\n A. 532 v\u00e0 329 (tri\u1ec7u \u0111\u1ed3ng) \u00a0\u00a0\u00a0C. 600 v\u00e0 292 (tri\u1ec7u \u0111\u1ed3ng)<\/p>\n B. 532 v\u00e0 292 (tri\u1ec7u \u0111\u1ed3ng) \u00a0\u00a0\u00a0D. 600 v\u00e0 329 (tri\u1ec7u \u0111\u1ed3ng)<\/p>\n C\u00e2u 9:<\/b>\u00a0T\u00ecm \u0111\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 y = x.23x<\/sup><\/p>\n A. y’ = 23x<\/sup>(1 + 3xln2) \u00a0\u00a0\u00a0C. y’ = 23x<\/sup>(1 + 3ln3)<\/p>\n B. y’ = 23x<\/sup>(1 + xln2) \u00a0\u00a0\u00a0D. y’ = 23x<\/sup>(1 + xln3)<\/p>\n C\u00e2u 10:<\/b>\u00a0T\u00ednh \u0111\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1<\/p>\n <\/p>\n C\u00e2u 11:<\/b>\u00a0T\u00ecm \u0111\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1<\/p>\n <\/p>\n C\u00e2u 12:<\/b>\u00a0Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = xe-2x<\/sup>\u00a0+ 2 t\u1ea1i giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 v\u1edbi tr\u1ee5c tung<\/p>\n A. y = x + 2 \u00a0\u00a0\u00a0B. y = x \u00a0\u00a0\u00a0C. y = 2x + 2 \u00a0\u00a0\u00a0D. y = -2x + 2<\/p>\n C\u00e2u 13:<\/b>\u00a0T\u00ecm c\u00e1c kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = 4x – 5ln(x2<\/sup>\u00a0+ 1)<\/p>\n <\/p>\n C\u00e2u 14:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x2<\/sup>e-x<\/sup>\u00a0. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang ?<\/p>\n A. H\u00e0m s\u1ed1 c\u00f3 x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i, x = 2 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u<\/p>\n B. H\u00e0m s\u1ed1 c\u00f3 x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u, x = -2 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i<\/p>\n C. H\u00e0m s\u1ed1 c\u00f3 x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i, x = -2 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u<\/p>\n D. H\u00e0m s\u1ed1 c\u00f3 x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u, x = 2 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 8:<\/b><\/p>\n Gi\u00e1 tr\u1ecb xe l\u00fac mua: G(0) = 600 tri\u1ec7u \u0111\u1ed3ng<\/p>\n Gi\u00e1 tr\u1ecb xe sau khi mua 5 n\u0103m : G(5) = 600.e-0,12.5<\/sup>\u00a0\u2248 329 tri\u1ec7u \u0111\u1ed3ng<\/p>\n C\u00e2u 9:<\/b><\/p>\n y’ = 23x<\/sup>\u00a0+ x.23x<\/sup>.ln(2)3 = 23x<\/sup>(1 + 3xln2)<\/p>\n C\u00e2u 10:<\/b><\/p>\n <\/p>\n C\u00e2u 11:<\/b><\/p>\n \u0110\u1ec3 thu\u1eadn ti\u1ec7n, ta vi\u1ebft l\u1ea1i<\/p>\n <\/p>\n C\u00e2u 12:<\/b><\/p>\n y’ = e-2x<\/sup>(1 – 2x); y'(0) = 1, y(0) = 2. Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ea7n t\u00ecm : y = x + 2<\/p>\n C\u00e2u 13:<\/b><\/p>\n T\u1eadp x\u00e1c \u0111\u1ecbnh : R<\/p>\n <\/p>\n B\u1ea3ng x\u00e9t d\u1ea5u<\/p>\n <\/p>\n Kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 (-\u221e; 1\/2) v\u00e0 (2; +\u221e)<\/p>\n C\u00e2u 14:<\/b><\/p>\n y’ = e-x<\/sup>x(2 – x). B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n T\u1eeb b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y x = 0 l\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u, x = 2 l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 8:\u00a0Gi\u00e1 tr\u1ecb c\u1ee7a m\u1ed9t chi\u1ebfc xe \u00f4 t\u00f4 sau t n\u0103m k\u1ec3 t\u1eeb khi mua \u0111\u01b0\u1ee3c \u01b0\u1edbc l\u01b0\u1ee3ng b\u1eb1ng c\u00f4ng th\u1ee9c G(t) = 600e-0,12t\u00a0(tri\u1ec7u \u0111\u1ed3ng). T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a chi\u1ebfc xe n\u00e0y t\u1ea1i hai th\u1eddi \u0111i\u1ec3m : l\u00fac mua v\u00e0 l\u00fac \u0111\u00e3 s\u1eed d\u1ee5ng 5 n\u0103m (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng tri\u1ec7u) A. […]<\/p>\n","protected":false},"author":3,"featured_media":27426,"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 8-C<\/td>\n 9-A<\/td>\n 10-C<\/td>\n 11-B<\/td>\n 12-A<\/td>\n 13-C<\/td>\n 14-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n