C\u00e2u 7:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 2x2<\/sup>\u00a0+ mx – 1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb?<\/p>\n <\/p>\n C\u00e2u 8:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, h\u00e0m s\u1ed1 y = -mx4<\/sup>\u00a0+ 2(m – 1)x2<\/sup>\u00a0+ 1 – 2m c\u00f3 m\u1ed9t c\u1ef1c tr\u1ecb<\/p>\n A.0 \u2264 m \u2264 1 \u00a0\u00a0\u00a0 B. m > 1 ho\u1eb7c m < 0 \u00a0\u00a0\u00a0 C. 0 < m < 1 \u00a0\u00a0\u00a0 D. 0 < m \u2264 1<\/p>\n C\u00e2u 9:<\/b>\u00a0Gi\u00e1 tr\u1ecb c\u1ee7a m \u0111\u1ec3 h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 3mx2<\/sup>\u00a0+ (m2<\/sup>\u00a0– 1)x + 2 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 2 l\u00e0:<\/p>\n A. m = 1 \u00a0\u00a0\u00a0 B. m = 11 \u00a0\u00a0\u00a0 C. m = -1 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n C\u00e2u 10:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, h\u00e0m s\u1ed1 y = (x – m)3<\/sup>\u00a0– 3x \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 x = 0?<\/p>\n A. m = 1 \u00a0\u00a0\u00a0B. m = -1 \u00a0\u00a0\u00a0 C. m = 0 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n C\u00e2u 11:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0+ 2(m – 1)x2<\/sup>\u00a0+ (m2<\/sup>\u00a0– 4m + 1)x + 2(m2<\/sup>\u00a0+ 1) c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb x1<\/sub>,x2<\/sub>\u00a0th\u1ecfa m\u00e3n<\/p>\n <\/p>\n A. m = 1\/2 \u00a0\u00a0\u00a0B. m = 2 \u00a0\u00a0\u00a0 C. m = 1\/2 ho\u1eb7c m = 2 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n C\u00e2u 12:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– mx2<\/sup>\u00a0+ 3(m2<\/sup>\u00a0– 1)x – m\u00a03<\/sup>\u00a0+ m c\u00f3 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i B, \u0111i\u1ec3m c\u1ef1c ti\u1ec3u C th\u1ecfa m\u00e3n OC = 3OB, v\u1edbi O l\u00e0 g\u1ed1c t\u1ecda \u0111\u1ed9?<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 7:<\/b><\/p>\n y’ = 3x2<\/sup>\u00a0– 4x + m. H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb <=> y\u2019=0 v\u00f4 nghi\u1ec7m ho\u1eb7c c\u00f3 nghi\u1ec7m k\u00e9p <=> \u0394’ \u2264 0 <=> 22<\/sup>\u00a0– 3m \u2264 0 <=> m \u2265 4\/3<\/p>\n Do \u0111\u00f3 h\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb khi m \u2265 4\/3<\/p>\n C\u00e2u 8:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1 y = -mx4<\/sup>\u00a0+2(m – 1)x2<\/sup>\u00a0+ 1 – 2m(1)<\/p>\n TH1: m = 0 (1) tr\u1edf th\u00e0nh y = -2x2<\/sup>\u00a0+ 1<\/p>\n V\u1eady v\u1edbi m = 0 h\u00e0m s\u1ed1 lu\u00f4n c\u00f3 m\u1ed9t c\u1ef1c tr\u1ecb.<\/p>\n TH2: m \u2260 0. y’ = -4mx3<\/sup>\u00a0+ 4(m – 1)x<\/p>\n <\/p>\n \u0110\u1ec3 h\u00e0m s\u1ed1 (1) c\u00f3 m\u1ed9t c\u1ef1c tr\u1ecb th\u00ec<\/p>\n <\/p>\n v\u00f4 nghi\u1ec7m ho\u1eb7c c\u00f3 nghi\u1ec7m k\u00e9p b\u1eb1ng 0<\/p>\n <\/p>\n K\u1ebft h\u1ee3p c\u1ea3 hai tr\u01b0\u1eddng h\u1ee3p ta c\u00f3 0 \u2264 m \u2264 1<\/p>\n C\u00e2u 9:<\/b><\/p>\n y’ = 3x2<\/sup>\u00a0– 6mx + m2<\/sup>\u00a0– 1; y” = 6x – 6m<\/p>\n H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 2 khi<\/p>\n <\/p>\n C\u00e2u 10:<\/b><\/p>\n X\u00e9t y = x3<\/sup>\u00a0– 3mx2<\/sup>\u00a0+ (3m2<\/sup>\u00a0– 3)x – m2<\/sup><\/p>\n Ta c\u00f3: y’ = 32<\/sup>\u00a0– 6mx + 3m2<\/sup>\u00a0– 3, y” = 6x – 6m<\/p>\n H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 x = 0 khi<\/p>\n <\/p>\n C\u00e2u 11:<\/b><\/p>\n Ta c\u00f3 y’ = 3x2<\/sup>\u00a0+ 4(m – 1)x + m2<\/sup>\u00a0– 4m + 1. H\u00e0m s\u1ed1 c\u00f3 hai c\u1ef1c tr\u1ecb<\/p>\n => y’ = 0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <=> \u0394’ > 0 <=> 4(m – 1)2<\/sup>\u00a0– 3(m2<\/sup>\u00a0– 4m + 1) > 0<\/p>\n <=> m2<\/sup>\u00a0+ 4m + 1 > 0<\/p>\n <\/p>\n \u00c1p d\u1ee5ng Vi-\u00e9t cho ph\u01b0\u01a1ng tr\u00ecnh y\u2019 = 0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1<\/sub>, x2<\/sub>\u00a0ta c\u00f3 :<\/p>\n <\/p>\n <\/p>\n \u0110\u1ed1i chi\u1ebfu \u0111i\u1ec1u ki\u1ec7n (*) c\u00f3 m = 5 ho\u1eb7c m = 1<\/p>\n C\u00e2u 12:<\/b><\/p>\n Ta c\u00f3 y’ = 3x2<\/sup>\u00a0– 6mx + 3(m2<\/sup>\u00a0– 1).<\/p>\n H\u00e0m s\u1ed1 c\u00f3 hai c\u1ef1c tr\u1ecb => y’ = 0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <=> \u0394’ > 0 <=> (3m)2<\/sup>\u00a0– 3.3(m2<\/sup>\u00a0– 1) > 0 <=> 9 > 0 \u0111\u00fang v\u1edbi m\u1ecdi m. Ta c\u00f3 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i l\u00e0 B(m – 1; -2m + 2) v\u00e0 c\u1ef1c ti\u1ec3u l\u00e0 C(m + 1; -2m – 2)<\/p>\n <\/p>\n <=> 40m2<\/sup>\u00a0– 100m + 40 = 0<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 7:\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, h\u00e0m s\u1ed1 y = x3\u00a0– 2×2\u00a0+ mx – 1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb? C\u00e2u 8:\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, h\u00e0m s\u1ed1 y = -mx4\u00a0+ 2(m – 1)x2\u00a0+ 1 – 2m c\u00f3 m\u1ed9t c\u1ef1c tr\u1ecb A.0 \u2264 m \u2264 1 \u00a0\u00a0\u00a0 B. m > 1 ho\u1eb7c m < […]<\/p>\n","protected":false},"author":3,"featured_media":28042,"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 7-A<\/td>\n 8-A<\/td>\n 9-B<\/td>\n 10-B<\/td>\n 11-C<\/td>\n 12-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n