C\u00e2u 19:<\/b>\u00a0Trong c\u00e1c ti\u1ebfp tuy\u1ebfn t\u1ea1i c\u00e1c \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 2, ti\u1ebfp tuy\u1ebfn c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c nh\u1ecf nh\u1ea5t b\u1eb1ng:<\/p>\n A. -3\u00a0\u00a0\u00a0B. 3\u00a0\u00a0\u00a0C. -4\u00a0\u00a0\u00a0D. 0<\/p>\n C\u00e2u 20:<\/b>\u00a0H\u00e0m s\u1ed1 n\u00e0o sau \u0111\u00e2y c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean nh\u01b0 h\u00ecnh b\u00ean?<\/p>\n <\/p>\n <\/p>\n C\u00e2u 21:<\/b>\u00a0H\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ mx \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 2 khi:<\/p>\n A.m < 0\u00a0\u00a0\u00a0B. m > 0\u00a0\u00a0\u00a0 C. m = 0 \u00a0\u00a0\u00a0D. m \u2260 0<\/p>\n C\u00e2u 22:<\/b>\u00a0H\u00e0m s\u1ed1<\/p>\n <\/p>\n \u0111\u1ed3ng bi\u1ebfn tr\u00ean t\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a n\u00f3 khi:<\/p>\n A. -2 \u2264 m \u2264 -1 \u00a0\u00a0\u00a0B. -2 < m < -1\u00a0\u00a0\u00a0 C. m < -2\u00a0\u00a0\u00a0D. m > -1<\/p>\n C\u00e2u 23:<\/b>\u00a0Cho \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 2x (C). G\u1ecdi x1<\/sub>,x2<\/sub>\u00a0l\u00e0 ho\u00e0nh \u0111\u1ed9 c\u00e1c \u0111i\u1ec3m M, N tr\u00ean (C), m\u00e0 t\u1ea1i \u0111\u00f3 ti\u1ebfp tuy\u1ebfn c\u1ee7a (C) vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng y = -x + 2017 . Khi \u0111\u00f3 (x1<\/sub>, x2<\/sub>) b\u1eb1ng<\/p>\n A. 4\u00a0\u00a0\u00a0B. -4\/3 \u00a0\u00a0\u00a0C. 4\/3 \u00a0\u00a0\u00a0 D. -1<\/p>\n C\u00e2u 24:<\/b>\u00a0M\u1ed9t ng\u1ecdn h\u1ea3i \u0111\u0103ng \u0111\u1eb7t tr\u1ea1i v\u1ecb tr\u00ed A c\u00e1ch b\u1eddbi\u1ec3n m\u1ed9t kho\u1ea3ng AB = 5km. Tr\u00ean b\u1edd bi\u1ec3n c\u00f3 m\u1ed9t kho v\u1ecb tr\u00ed C c\u00e1ch B m\u1ed9t kho\u1ea3ng l\u00e0 7km. Do \u0111\u1ecba h\u00ecnh hi\u1ec3m tr\u1edf, ng\u01b0\u1eddi canh h\u1ea3i \u0111\u0103ng ch\u1ec9 c\u00f3 th\u1ec3 ch\u00e8o thuy\u1ec1n t\u1eeb A \u0111\u1ebfn M tr\u00ean b\u1edd bi\u1ec3n v\u1edbi v\u1eadn t\u1ed1c 4km\/h r\u1ed3i \u0111i b\u1ed9 \u0111\u1ebfn C, v\u1edbi v\u1eadn t\u1ed1c 6km\/h. V\u1eady v\u1ecb tr\u00ed M c\u00e1ch B m\u1ed9t kho\u1ea3ng bao xa th\u00ec ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn kho l\u00e0 nhanh nh\u1ea5t?<\/p>\n <\/p>\n A. 3,5km\u00a0\u00a0\u00a0B. 4,5km\u00a0\u00a0\u00a0C. 5,5km\u00a0\u00a0\u00a0D. 6,5km<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 19:<\/b><\/p>\n Ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c l\u00e0<\/p>\n k = y’ = 3x2<\/sup>\u00a0– 6x = 3(x – 1)2<\/sup>\u00a0– 3 > -3 \u2200x \u2208 R<\/p>\n C\u00e2u 21:<\/b><\/p>\n H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x=2<\/p>\n C\u00e2u 23:<\/b><\/p>\n Ti\u1ebfp tuy\u1ebfn c\u1ee7a C vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng y= -x + 2017 n\u00ean h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a ti\u1ebfp tuy\u1ebfn l\u00e0 k2<\/sub>\u00a0th\u1ecfa m\u00e3n (-1)k2<\/sub>\u00a0= -1 => k2<\/sub>\u00a0= 1<\/p>\n Suy ra k2<\/sub>\u00a0= y’ = 1 => 3x2<\/sup>\u00a0– 4x + 2 <=> 3x2<\/sup>\u00a0– 4x + 2 = 0 (*)<\/p>\n V\u00ec x1<\/sub>, x2<\/sub>\u00a0l\u00e0 nghi\u1ec7m c\u1ee7a (*) n\u00ean \u00e1p d\u1ee5ng Vi-\u00e9t ta c\u00f3 x1<\/sub>\u00a0+ x2<\/sub>\u00a0= 4\/3<\/p>\n C\u00e2u 24:<\/b><\/p>\n \u0110\u1eb7t BM = x (0 \u2264 x \u2264 7) => MC = 7 – x. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py-ta-go cho tam gi\u00e1c vu\u00f4ng ABM c\u00f3<\/p>\n <\/p>\n Th\u1eddi gian \u0111i t\u1eeb A \u0111\u1ebfn M l\u00e0<\/p>\n <\/p>\n th\u1eddi gian \u0111i t\u1eeb M \u0111\u1ebfn C l\u00e0<\/p>\n <\/p>\n T\u1ed5ng th\u1eddi gian \u0111i t\u1eeb A \u0111\u1ebfn C l\u00e0<\/p>\n <\/p>\n <\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n \u0110\u1ec3 ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn kho nhanh nh\u1ea5t th\u00ec th\u1eddi gian \u0111i c\u1ea7n \u00edt nh\u1ea5t, t\u1ee9c t \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. D\u1ef1a v\u00e0o b\u1ea3ng bi\u1ebfn thi\u00ean ta th\u1ea5y t \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t t\u1ea1i x = 2\u221a5 \u2248 4,5<\/p>\n V\u1eady v\u1ecb tr\u00ed \u0111i\u1ec3m M c\u00e1ch B m\u1ed9t kho\u1ea3ng l\u00e0 4,5km th\u00ec ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn kho l\u00e0 nhanh nh\u1ea5t.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 19:\u00a0Trong c\u00e1c ti\u1ebfp tuy\u1ebfn t\u1ea1i c\u00e1c \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3\u00a0– 3×2\u00a0+ 2, ti\u1ebfp tuy\u1ebfn c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c nh\u1ecf nh\u1ea5t b\u1eb1ng: A. -3\u00a0\u00a0\u00a0B. 3\u00a0\u00a0\u00a0C. -4\u00a0\u00a0\u00a0D. 0 C\u00e2u 20:\u00a0H\u00e0m s\u1ed1 n\u00e0o sau \u0111\u00e2y c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean nh\u01b0 h\u00ecnh b\u00ean? C\u00e2u 21:\u00a0H\u00e0m s\u1ed1 y = x3\u00a0– 3×2\u00a0+ mx \u0111\u1ea1t c\u1ef1c […]<\/p>\n","protected":false},"author":3,"featured_media":27701,"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 19-A<\/td>\n 20-A<\/td>\n 21-C<\/td>\n 22-A<\/td>\n 23-C<\/td>\n 24-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n