C\u00e2u 1:<\/b>\u00a0\u0110\u01b0\u1eddng cong trong h\u00ecnh d\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a m\u1ed9t h\u00e0m s\u1ed1 trong 4 h\u00e0m s\u1ed1 \u0111\u01b0\u1ee3c li\u1ec7t k\u00ea \u1edf 4 ph\u01b0\u01a1ng \u00e1n A, B, C, D d\u01b0\u1edbi \u0111\u00e2y. H\u1ecfi h\u00e0m s\u1ed1 \u0111\u00f3 l\u00e0 h\u00e0m s\u1ed1 n\u00e0o?<\/p>\n
<\/p>\n
A. y = x4<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 2<\/p>\n B. y = x3<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 1<\/p>\n C. y = -4x4<\/sup>\u00a0+ x2<\/sup>\u00a0+ 4<\/p>\n D. y = x4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 3<\/p>\n C\u00e2u 2:<\/b>\u00a0\u0110\u1ed3 th\u1ecb trong h\u00ecnh d\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 n\u00e0o?<\/p>\n <\/p>\n A. y = x2<\/sup>\u00a0– 2x + 1<\/p>\n B. y = x3<\/sup>\u00a0+ 4x2<\/sup>\u00a0– 2x + 5<\/p>\n C. y = x4<\/sup>\u00a0+ x2<\/sup>\u00a0+ 1<\/p>\n D. y = x4<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 5<\/p>\n C\u00e2u 3:<\/b>\u00a0T\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = -x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 1 l\u00e0:<\/p>\n A. (-1; -1) \u00a0\u00a0\u00a0 B. (-2; -3) \u00a0\u00a0\u00a0 C. (0; 1) \u00a0\u00a0\u00a0 D. Kh\u00f4ng c\u00f3 \u0111\u00e1p \u00e1n<\/p>\n C\u00e2u 4:<\/b>\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau, m\u1ec7nh \u0111\u1ec1 n\u00e0o \u0111\u00fang?<\/p>\n A. I(1; 0) l\u00e0 t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a<\/p>\n <\/p>\n B. I(1; 0) l\u00e0 t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a y = -x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 2<\/p>\n C. I(1; 0) l\u00e0 \u0111i\u1ec3m thu\u1ed9c \u0111\u1ed3 th\u1ecb<\/p>\n <\/p>\n D. I(1; 0) l\u00e0 giao \u0111i\u1ec3m c\u1ee7a y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0– 2 v\u1edbi tr\u1ee5c ho\u00e0nh.<\/p>\n C\u00e2u 5:<\/b>\u00a0T\u00ecm m \u0111\u1ec3 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh x4<\/sup>\u00a0+ 2x2<\/sup>\u00a0\u2265 m lu\u00f4n \u0111\u00fang.<\/p>\n A. m = 0 \u00a0\u00a0\u00a0 B. m < 0<\/p>\n C. m \u2264 0 \u00a0\u00a0\u00a0 D. Kh\u00f4ng c\u00f3 \u0111\u00e1p \u00e1n<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n Theo Ch\u00fa \u00fd \u1edf m\u1ee5c 2, \u0111\u1ed3 th\u1ecb \u1ee9ng v\u1edbi h\u00e0m b\u1eadc b\u1ed1n tr\u00f9ng ph\u01b0\u01a1ng c\u00f3 a > 0 v\u00e0 a, b, tr\u00e1i d\u1ea5u.<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n D.<\/p>\n C\u00e2u 2:<\/b><\/p>\n Theo Ch\u00fa \u00fd \u1edf m\u1ee5c 2, \u0111\u1ed3 th\u1ecb \u1ee9ng v\u1edbi h\u00e0m b\u1eadc b\u1ed1n tr\u00f9ng ph\u01b0\u01a1ng c\u00f3 a > 0 v\u00e0 a, b, c\u00f9ng d\u1ea5u ho\u1eb7c h\u00e0m s\u1ed1 b\u1eadc hai v\u1edbi a > 0. Tuy nhi\u00ean \u0111\u1ec9nh c\u1ee7a parabol c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = -x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 1 l\u00e0 I(1; 0) n\u1eb1m tr\u00ean tr\u1ee5c ho\u00e0nh.<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n C\u00e2u 3:<\/b><\/p>\n y’ = -3x2<\/sup>\u00a0– 6x; y” = -6x – 6; y” = 0 => x = -1<\/p>\n V\u1eady \u0111i\u1ec3m U(-1; -1) l\u00e0 t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (theo ch\u00fa \u00fd \u1edf m\u1ee5c 1).<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n C\u00e2u 4:<\/b><\/p>\n A. T\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a<\/p>\n <\/p>\n C. \u0110i\u1ec3m I(1; 0) kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb<\/p>\n <\/p>\n D. \u0110i\u1ec3m I(1; 0) kh\u00f4ng thu\u1ed9c \u0111\u1ed3 th\u1ecb y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0– 2 n\u00ean kh\u00f4ng ph\u1ea3i l\u00e0 giao \u0111i\u1ec3m c\u1ee7a y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0– 2 v\u1edbi tr\u1ee5c ho\u00e0nh.<\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n B.<\/p>\n C\u00e2u 5:<\/b><\/p>\n X\u00e9t h\u00e0m y = x4<\/sup>\u00a0+ 2x2<\/sup>\u00a0\u2264 m c\u00f3 a = 1 > 0; b = 2 > 0 => a, b c\u00f9ng d\u1ea5u.<\/p>\n \u0110\u1ed3 th\u1ecb c\u00f3 d\u1ea1ng nh\u01b0 h\u00ecnh b\u00ean.<\/p>\n T\u1eeb \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 ta suy ra m \u2264 0 . Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0\u0110\u01b0\u1eddng cong trong h\u00ecnh d\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a m\u1ed9t h\u00e0m s\u1ed1 trong 4 h\u00e0m s\u1ed1 \u0111\u01b0\u1ee3c li\u1ec7t k\u00ea \u1edf 4 ph\u01b0\u01a1ng \u00e1n A, B, C, D d\u01b0\u1edbi \u0111\u00e2y. H\u1ecfi h\u00e0m s\u1ed1 \u0111\u00f3 l\u00e0 h\u00e0m s\u1ed1 n\u00e0o? A. y = x4\u00a0+ 3×2\u00a0– 2 B. y = x3\u00a0– 2×2\u00a0+ 1 C. y = -4×4\u00a0+ […]<\/p>\n","protected":false},"author":3,"featured_media":27904,"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-D<\/td>\n 2-C<\/td>\n 3-A<\/td>\n 4-B<\/td>\n 5-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n