C\u00e2u 1:<\/b>\u00a0Cho h\u00e0m s\u1ed1 f c\u00f3 \u0111\u1ea1o h\u00e0m l\u00e0 f'(x) = x(x+1)2<\/sup>(x-2)4<\/sup>\u00a0v\u1edbi m\u1ecdi x \u2208 R. S\u1ed1 \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 f l\u00e0:<\/p>\n A. 0 \u00a0\u00a0\u00a0 B. 1 \u00a0\u00a0\u00a0 C. 2 \u00a0\u00a0\u00a0 D.3<\/p>\n C\u00e2u 2:<\/b>\u00a0\u0110i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 y = -x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 1 l\u00e0:<\/p>\n A. x = 0 \u00a0\u00a0\u00a0 B. x = -2 \u00a0\u00a0\u00a0 C. x = 2 \u00a0\u00a0\u00a0D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n C\u00e2u 3:<\/b>\u00a0\u0110i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0+ 4x2<\/sup>\u00a0+ 2 l\u00e0:<\/p>\n A. x = 1 \u00a0\u00a0\u00a0 B. x = \u221a2 \u00a0\u00a0\u00a0 C. x = 0 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n C\u00e2u 4:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 2x2<\/sup>\u00a0– 1 (1) v\u00e0 c\u00e1c m\u1ec7nh \u0111\u1ec1<\/p>\n (1) \u0110i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 (1) l\u00e0 x = 0 ho\u1eb7c x = 4\/3<\/p>\n (2) \u0110i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 (1) l\u00e0 x = 0 v\u00e0 x = 4\/3<\/p>\n (3) \u0110i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (1) l\u00e0 x = 0 v\u00e0 x = 4\/3<\/p>\n (4) C\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 (1) l\u00e0 x = 0 v\u00e0 x = 4\/3<\/p>\n Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 tr\u00ean, s\u1ed1 m\u1ec7nh \u0111\u1ec1 sai l\u00e0:<\/p>\n A.0 \u00a0\u00a0\u00a0 B.1 \u00a0\u00a0\u00a0 C.2 \u00a0\u00a0\u00a0 D.3<\/p>\n C\u00e2u 5:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 2x2<\/sup>\u00a0– 2 (2). Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\n A. H\u00e0m s\u1ed1 (2) \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i y = -2<\/p>\n B. H\u00e0m s\u1ed1 (2) \u0111\u1ea1t gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i t\u1ea1i y = -2<\/p>\n C. \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (2) c\u00f3 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i l\u00e0 y = -2<\/p>\n D. H\u00e0m s\u1ed1 (2) c\u00f3 gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i l\u00e0 y = -2<\/p>\n C\u00e2u 6:<\/b>\u00a0H\u00e0m s\u1ed1 y = cosx \u0111\u1ea1t c\u1ef1c tr\u1ecb t\u1ea1i nh\u1eefng \u0111i\u1ec3m<\/p>\n <\/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<\/p>\n <\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n D\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 0. V\u1eady h\u00e0m s\u1ed1 c\u00f3 m\u1ed9t c\u1ef1c tr\u1ecb<\/p>\n C\u00e2u 2:<\/b><\/p>\n Ta c\u00f3 y’ = -3x2<\/sup>\u00a0– 6x, y” = -6x – 6 .<\/p>\n X\u00e9t<\/p>\n <\/p>\n y”(0) = -6 < 0; y”(-2) = 6 > 0<\/p>\n Do \u0111\u00f3 h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 0<\/p>\n C\u00e2u 3:<\/b><\/p>\n Ta c\u00f3: y’ = 4x3<\/sup>\u00a0+ 8x, y” = 12x2<\/sup>\u00a0+ 8. y’ = 0 <=> 4x(x2<\/sup>\u00a0+ 2) = 0 <=> x = 0<\/p>\n y”(0) = 2 > 0. Do \u0111\u00f3 h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 0<\/p>\n C\u00e2u 4:<\/b><\/p>\n Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 4x, y” = 6x – 4;<\/p>\n <\/p>\n y”(0) = -4 < 0; y”(4\/3) = 4 > 0. Do \u0111\u00f3 h\u00e0m s\u1ed1 c\u00f3 hai c\u1ef1c tr\u1ecb l\u00e0 x = 0 v\u00e0 x = 4\/3<\/p>\n C\u00e2u 5:<\/b><\/p>\n Ta c\u00f3: y’ = 4x3<\/sup>\u00a0– 4x, y” = 12x2<\/sup>\u00a0– 4<\/p>\n <\/p>\n y”(-1) = 8 > 0; y”(1) = 8 > 0<\/p>\n Do \u0111\u00f3 h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 0 v\u00e0 c\u00f3 gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i l\u00e0 y(0)=-2<\/p>\n C\u00e2u 6:<\/b><\/p>\n y’ = -sinx; y” = -cosx. y’ = 0 <=> -sinx = 0 <=> x = k\u03c0<\/p>\n y”(k\u03c0) = \u00b11. Do \u0111\u00f3 h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c tr\u1ecb t\u1ea1i x = k\u03c0<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Cho h\u00e0m s\u1ed1 f c\u00f3 \u0111\u1ea1o h\u00e0m l\u00e0 f'(x) = x(x+1)2(x-2)4\u00a0v\u1edbi m\u1ecdi x \u2208 R. S\u1ed1 \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 f l\u00e0: A. 0 \u00a0\u00a0\u00a0 B. 1 \u00a0\u00a0\u00a0 C. 2 \u00a0\u00a0\u00a0 D.3 C\u00e2u 2:\u00a0\u0110i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 y = -x3\u00a0– 3×2\u00a0+ 1 l\u00e0: A. x = 0 \u00a0\u00a0\u00a0 B. x […]<\/p>\n","protected":false},"author":3,"featured_media":28056,"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-A<\/td>\n 3-C<\/td>\n 4-D<\/td>\n 5-D<\/td>\n 6-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n