C\u00e2u 7:<\/b>\u00a0H\u00e0m s\u1ed1:<\/p>\n
<\/p>\n
\u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng n\u00e0o?<\/p>\n
A. R \u00a0\u00a0\u00a0B. (-\u221e; 0) \u00a0\u00a0\u00a0 C. (-1; 0) \u00a0\u00a0\u00a0D. (0; +\u221e)<\/p>\n
C\u00e2u 8:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– x2<\/sup>\u00a0+ (m-1)x + m. T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R<\/p>\n A. m \u2264 2 \u00a0\u00a0\u00a0 B. m > 2 \u00a0\u00a0\u00a0 C. m \u2265 2 \u00a0\u00a0\u00a0D. m <2<\/p>\n C\u00e2u 9:<\/b>\u00a0Cho h\u00e0m s\u1ed1<\/p>\n <\/p>\n T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1).<\/p>\n A. m < 2\u221a2 \u00a0\u00a0\u00a0B. m \u2265 -2\u221a2 \u00a0\u00a0\u00a0 C. m = 2\u221a2 \u00a0\u00a0\u00a0 D. -2\u221a2 \u2264 m 2\u221a2<\/p>\n C\u00e2u 10:<\/b>\u00a0T\u00ecm t\u1ea5t c\u1ea3 c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 m sao cho h\u00e0m s\u1ed1<\/p>\n <\/p>\n A. 1 < m < 5 \u00a0\u00a0\u00a0B. m \u2265 5 \u00a0\u00a0\u00a0 C. m < -1 ho\u1eb7c m > 5 \u00a0\u00a0\u00a0 D. m > 5<\/p>\n C\u00e2u 11:<\/b>\u00a011. Cho h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ mx + 1 – 2m. T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a m \u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean \u0111o\u1ea1n c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng 1.<\/p>\n A. m =0 \u00a0\u00a0\u00a0 B. m = 1\/4 \u00a0\u00a0\u00a0 C. 9\/4 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 7:<\/b><\/p>\n <\/p>\n H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R<\/p>\n C\u00e2u 8:<\/b><\/p>\n y’ = x2<\/sup>\u00a0– 2x + (m -1). H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R<\/p>\n <=> y’ > 0 \u2200x \u2208 R <=> \u0394’ \u2265 0; \u0394’ = -m + 2 \u2265 0 <=> m > 2<\/p>\n C\u00e2u 9:<\/b><\/p>\n Ta c\u00f3 y’ = -x2<\/sup>\u00a0– mx – 2 . H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; – 1) n\u1ebfu y’ = x2<\/sup>\u00a0– mx – 2 \u2264 0 tr\u00ean kho\u1ea3ng (-\u221e; -1)<\/p>\n C\u00e1ch 1. D\u00f9ng \u0111\u1ecbnh l\u00ed d\u1ea5u c\u1ee7a tam th\u1ee9c b\u1eadc hai. Ta c\u00f3 \u0394 = m2<\/sup>\u00a0– 8<\/p>\n TH1: -2\u221a2 \u2264 m \u2264 2\u221a2 => \u0394 \u2264 0. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean R<\/p>\n TH2:<\/p>\n <\/p>\n y’ = 0. c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0<\/p>\n <\/p>\n T\u1eeb TH1 v\u00e0 TH2, ta c\u00f3 m \u2264 2\u221a2<\/p>\n C\u00e1ch 2. D\u00f9ng ph\u01b0\u01a1ng ph\u00e1p bi\u1ebfn thi\u00ean h\u00e0m s\u1ed1<\/p>\n Ta c\u00f3<\/p>\n <\/p>\n T\u1eeb \u0111\u00f3 suy ra<\/p>\n <\/p>\n Do \u0111\u00f3 m \u2264 2\u221a2<\/p>\n V\u1eady gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1) l\u00e0 m = 2\u221a2<\/p>\n C\u00e2u 10:<\/b><\/p>\n Ta c\u00f3<\/p>\n <\/p>\n C\u00e2u 11:<\/b><\/p>\n y’ = 3x2<\/sup>\u00a0+ 6x + m. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn n\u1ebfu y’ \u2265 0. Ta c\u00f3 \u0394’ = 9 – 3m<\/p>\n TH1: m \u2265 3 => \u0394’ \u2264 0 .<\/p>\n H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R. Do \u0111\u00f3 m \u2265 3 kh\u00f4ng th\u1ecfa m\u00e3n y\u00eau c\u1ea7u \u0111\u1ec1 b\u00e0i<\/p>\n TH2: m < 3 => \u0394’ > 0 .<\/p>\n y\u2019 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0<\/p>\n <\/p>\n T\u1eeb b\u1ea3ng bi\u1ebfn thi\u00ean, ta th\u1ea5y kh\u00f4ng t\u1ed3n t\u1ea1i m \u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean \u0111o\u1ea1n c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng 1.<\/p>\n T\u1eeb TH1 v\u00e0 TH2, kh\u00f4ng t\u1ed3n t\u1ea1i m th\u1ecfa m\u00e3n.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 7:\u00a0H\u00e0m s\u1ed1: \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng n\u00e0o? A. R \u00a0\u00a0\u00a0B. (-\u221e; 0) \u00a0\u00a0\u00a0 C. (-1; 0) \u00a0\u00a0\u00a0D. (0; +\u221e) C\u00e2u 8:\u00a0Cho h\u00e0m s\u1ed1 y = x3\u00a0– x2\u00a0+ (m-1)x + m. T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean R A. m \u2264 2 \u00a0\u00a0\u00a0 B. m > 2 \u00a0\u00a0\u00a0 […]<\/p>\n","protected":false},"author":3,"featured_media":28084,"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-C<\/td>\n 9-C<\/td>\n 10-D<\/td>\n 11-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n