C\u00e2u 13:<\/b>\u00a0Ha\u0300m s\u00f4\u0301<\/p>\n
<\/p>\n
\u0111a\u0323t gia\u0301 tri\u0323 l\u01a1\u0301n nh\u00e2\u0301t ta\u0323i x b\u0103\u0300ng<\/p>\n
A. 1 \u00a0\u00a0\u00a0B. 1\/2 \u00a0\u00a0\u00a0C. -2\u00a0\u00a0\u00a0D. -1.<\/p>\n
C\u00e2u 14:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x4<\/sup>\u00a0– 2m2<\/sup>x2<\/sup>\u00a0+ 1 co\u0301 ba c\u01b0\u0323c tri\u0323 ta\u0323o tha\u0300nh tam gia\u0301c vu\u00f4ng.<\/p>\n A. m = \u00b1 1\u00a0\u00a0\u00a0B. m = \u00b1 2 \u00a0\u00a0\u00a0 C. m = 3 \u00a0\u00a0\u00a0D. \u0110a\u0301p a\u0301n kha\u0301c.<\/p>\n C\u00e2u 15:<\/b>\u00a0Ti\u0301nh gia\u0301 tri\u0323 bi\u00ea\u0309u th\u01b0\u0301c log3<\/sub>5.log4<\/sub>9.log5<\/sub>2<\/p>\n A. 1\/2 \u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0C. 2\u00a0\u00a0\u00a0D. 3<\/p>\n C\u00e2u 16:<\/b>\u00a0Ti\u0300m \u0111a\u0323o ha\u0300m cu\u0309a ha\u0300m s\u00f4\u0301 y = (\u221a3)x2<\/sup><\/p>\n <\/p>\n C\u00e2u 17:<\/b>\u00a0N\u00ea\u0301u 4x<\/sup>\u00a0– 4x – 1<\/sup>\u00a0= 24 thi\u0300 (2x)x<\/sup>\u00a0b\u0103\u0300ng<\/p>\n A. 5\u221a5\u00a0\u00a0\u00a0B. 25\u00a0\u00a0\u00a0C. 25\u221a5\u00a0\u00a0\u00a0D. 125.<\/p>\n C\u00e2u 18:<\/b>\u00a0Gia\u0309i ph\u01b0\u01a1ng tri\u0300nh log3<\/sub>x + log9<\/sub>x + log81<\/sub>x = 7<\/p>\n A. x = 27\u00a0\u00a0\u00a0B. x = 81\u00a0\u00a0\u00a0C. x = 729 \u00a0\u00a0\u00a0D. x = 243<\/p>\n C\u00e2u 19:<\/b>\u00a0N\u00ea\u0301u (log3<\/sub>x)(log2x<\/sub>y) = logx<\/sub>x2<\/sup>\u00a0thi\u0300 y b\u0103\u0300ng<\/p>\n A. 9\u00a0\u00a0\u00a0B. 9\/2 \u00a0\u00a0\u00a0C. 18\u00a0\u00a0\u00a0D. 81<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 13:<\/b><\/p>\n <\/p>\n y’ = 0 => x = \u221a2; y(\u221a2) = 2\u221a2<\/p>\n <\/p>\n C\u00e2u 15:<\/b><\/p>\n log3<\/sub>5. log4<\/sub>9. log5<\/sub>2 = (log3<\/sub>5.log5<\/sub>2).log22<\/sup><\/sub>32<\/sup>\u00a0= log3<\/sub>2.log2<\/sub>3 = 1<\/p>\n C\u00e2u 16:<\/b><\/p>\n y’ = (\u221a3)x2<\/sup>.ln\u221a3(x2<\/sup>)’<\/p>\n <\/p>\n C\u00e2u 17:<\/b><\/p>\n 4x<\/sup>\u00a0– 4x – 1<\/sup>\u00a0= 24 => 4x<\/sup>\u00a0– 4x<\/sup>\/4 = 24 => 3.4x<\/sup>\/4 = 24 => 22x<\/sup>\u00a0= 32 = 25<\/sup>\u00a0=> 2x = 5<\/p>\n => x = 5\/2 => (2x)x<\/sup>\u00a0= 55\/2<\/sup>\u00a0= 25\u221a5<\/p>\n C\u00e2u 18:<\/b><\/p>\n log3<\/sub>x + log9<\/sub>x + log81<\/sub>x = 7<\/p>\n <\/p>\n <=> log3<\/sub>x = 4 <=> x = 34<\/sup>\u00a0= 81<\/p>\n C\u00e2u 19:<\/b><\/p>\n (log3<\/sub>x)(logx<\/sub>(2x))(log2x<\/sub>y) = logx<\/sub>x2<\/sup>\u00a0<=> (log3<\/sub>(2x))log2x<\/sub>y = 2logx<\/sub>x <=> log3<\/sub>y = 2 <=> y = 32<\/sup>\u00a0= 9<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 13:\u00a0Ha\u0300m s\u00f4\u0301 \u0111a\u0323t gia\u0301 tri\u0323 l\u01a1\u0301n nh\u00e2\u0301t ta\u0323i x b\u0103\u0300ng A. 1 \u00a0\u00a0\u00a0B. 1\/2 \u00a0\u00a0\u00a0C. -2\u00a0\u00a0\u00a0D. -1. C\u00e2u 14:\u00a0Ti\u0300m m \u0111\u00ea\u0309 \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x4\u00a0– 2m2x2\u00a0+ 1 co\u0301 ba c\u01b0\u0323c tri\u0323 ta\u0323o tha\u0300nh tam gia\u0301c vu\u00f4ng. A. m = \u00b1 1\u00a0\u00a0\u00a0B. m = \u00b1 2 \u00a0\u00a0\u00a0 C. m = 3 \u00a0\u00a0\u00a0D. […]<\/p>\n","protected":false},"author":3,"featured_media":26649,"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 13-B<\/td>\n 14-A<\/td>\n 15-B<\/td>\n 16-B<\/td>\n 17-C<\/td>\n 18-B<\/td>\n 19-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n