C\u00e2u 7:<\/b>\u00a0Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x1\/5<\/sup>\u00a0t\u1ea1i \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 b\u1eb1ng 2.<\/p>\n <\/p>\n C\u00e2u 8:<\/b>\u00a0T\u00ednh t\u1ed5ng c\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n <\/p>\n A. 7.\u00a0\u00a0\u00a0B. 25. \u00a0\u00a0\u00a0C. 73. \u00a0\u00a0\u00a0 D.337.<\/p>\n C\u00e2u 9:<\/b>\u00a0T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb<\/p>\n <\/p>\n C\u00e2u 10:<\/b>\u00a0Cho 2 h\u00e0m s\u1ed1 f(x) = x2<\/sup>\u00a0v\u00e0 g(x) = x1\/2<\/sup>\u00a0. Bi\u1ebft r\u1eb1ng \u03b1 > 0 v\u00e0 f(\u03b1) < g(\u03b1). Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\n A. 0 < \u03b1 < 1\/2 \u00a0\u00a0\u00a0B. 0 < \u03b1 \u2264 1\u00a0\u00a0\u00a0C. 1\/2 < \u03b1 < 2\u00a0\u00a0\u00a0D. \u03b1 > 1<\/p>\n C\u00e2u 11:<\/b>\u00a0T\u00ecm c\u00e1c kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = \u221ax – \u221cx, x > 0<\/p>\n <\/p>\n C\u00e2u 12:<\/b>\u00a0T\u00ecm c\u00e1c \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1<\/p>\n <\/p>\n A. x=1. \u00a0\u00a0\u00a0B.x=2.<\/p>\n C. x=1 va x=-2\u00a0\u00a0\u00a0D. x=2 v\u00e0 x=-1.<\/p>\n C\u00e2u 13:<\/b>\u00a0T\u00ecm c\u00e1c \u0111i\u1ec3m c\u1ef1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1<\/p>\n <\/p>\n A. x=2.\u00a0\u00a0\u00a0B. 3\/2\u00a0\u00a0\u00a0C. x=6.\u00a0\u00a0\u00a0D. x=4.<\/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 Ti\u1ebfp tuy\u1ebfn<\/p>\n <\/p>\n C\u00e2u 8:<\/b><\/p>\n \u0110\u1eb7t t = \u221cx , nh\u1eadn \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh t2<\/sup>\u00a0– 7t + 12 = 0 <=> t = 3 ho\u1eb7c t = 4<\/p>\n => \u221cx = 3 ho\u1eb7c \u221cx = 4 => x = 34<\/sup>\u00a0= 81 ho\u1eb7c x = 44<\/sup>\u00a0= 256<\/p>\n T\u1ed5ng hai nghi\u1ec7m : 81 + 256 = 337<\/p>\n C\u00e2u 9:<\/b><\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m<\/p>\n <\/p>\n C\u00e2u 10:<\/b><\/p>\n <\/p>\n <=> \u03b13\/2 < 1 <=> 0 < \u03b1 < 1 (do \u03b1 > 0) => 0 < \u03b1 \u2264 1<\/p>\n C\u00e2u 11:<\/b><\/p>\n y’ = 0 <=> 2\u221cx – 1 > 0 <=> x > 1\/16 => Kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 (1\/16; +\u221e)<\/p>\n C\u00e2u 12:<\/b><\/p>\n <\/p>\n y’= 0 <=> x2<\/sup>\u00a0+ x – 2 = 0 <=> x = -2 (lo\u1ea1i) ho\u1eb7c x = 1<\/p>\n y’ \u0111\u1ed5i d\u1ea5u khi \u0111i qua \u0111i\u1ec3m x = 1 n\u00ean h\u00e0m s\u1ed1 c\u00f3 m\u1ed9t \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0 x = 1<\/p>\n C\u00e2u 13:<\/b><\/p>\n <\/p>\n y’ = 0 <=> 12 – 8x = 0 <=> x = 3\/2<\/p>\n y\u2019 \u0111\u1ed5i d\u1ea5u khi \u0111i qua \u0111i\u1ec3m x = 3\/2 n\u00ean h\u00e0m s\u1ed1 c\u00f3 m\u1ed9t \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0 x = 3\/2<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 7:\u00a0Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x1\/5\u00a0t\u1ea1i \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 b\u1eb1ng 2. C\u00e2u 8:\u00a0T\u00ednh t\u1ed5ng c\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh A. 7.\u00a0\u00a0\u00a0B. 25. \u00a0\u00a0\u00a0C. 73. \u00a0\u00a0\u00a0 D.337. C\u00e2u 9:\u00a0T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a \u0111\u1ed3 th\u1ecb C\u00e2u 10:\u00a0Cho 2 h\u00e0m s\u1ed1 f(x) = x2\u00a0v\u00e0 g(x) = x1\/2\u00a0. […]<\/p>\n","protected":false},"author":3,"featured_media":27564,"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-B<\/td>\n 8-D<\/td>\n 9-A<\/td>\n 10-B<\/td>\n 11-C<\/td>\n 12-A<\/td>\n 13-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n<\/p>\n