C\u00e2u 15:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301<\/p>\n
<\/p>\n
\u0110\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 co\u0301 ti\u00ea\u0323m c\u00e2\u0323n ngang la\u0300<\/p>\n
A. y = 1\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. y = 1\/2\u00a0\u00a0\u00a0D. y = -5<\/p>\n
C\u00e2u 16:<\/b>\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh |x3<\/sup>| – 12|x| = m (v\u01a1\u0301i -1 < m < 0 ) la\u0300<\/p>\n A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4.<\/p>\n C\u00e2u 17:<\/b>\u00a0Cho hai s\u00f4\u0301 d\u01b0\u01a1ng a, b tho\u0309a ma\u0303n a2<\/sup>\u00a0+ b2<\/sup>\u00a0= 7ab. \u0110\u0103\u0309ng th\u01b0\u0301c na\u0300o sau \u0111\u00e2y \u0111u\u0301ng?<\/p>\n <\/p>\n A. 1 \u00a0\u00a0\u00a0B. 2 \u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4.<\/p>\n C\u00e2u 18:<\/b>\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh log1\/4<\/sub>(x2<\/sup>\u00a0– x4<\/sup>) = 1 la\u0300<\/p>\n A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4.<\/p>\n C\u00e2u 19:<\/b>\u00a0Gia\u0309 s\u01b0\u0309 x la\u0300 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh:<\/p>\n <\/p>\n Khi \u0111o\u0301 ta co\u0301:<\/p>\n A. lg(1 – x) = 1\u00a0\u00a0\u00a0B. lg(1 – x) = \u221a3<\/p>\n C. lg(1 – x) < 1 \u00a0\u00a0\u00a0D. lg(1 – x) > \u221a3<\/p>\n C\u00e2u 20:<\/b>\u00a0T\u00e2\u0323p nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh 32x + 1<\/sup>\u00a0– 4.3x + 1<\/sup>\u00a0+ 9 \u2264 0 la\u0300<\/p>\n A. x \u2265 0\u00a0\u00a0\u00a0B. x \u2264 1 \u00a0\u00a0\u00a0 C. 0 \u2264 x \u2264 1 \u00a0\u00a0\u00a0D. 0 \u2264 x \u2264 2<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 15:<\/b><\/p>\n Ta c\u00f3<\/p>\n <\/p>\n Suy ra:<\/p>\n <\/p>\n V\u1eady \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 ti\u1ec7m c\u1eadn ngang l\u00e0 y = 1\/2<\/p>\n C\u00e2u 16:<\/b><\/p>\n S\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh |x3<\/sup>| – 12|x| = m l\u00e0 s\u1ed1 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d: y = m v\u1edbi \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = |x3<\/sup>| – 12|x|. D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb suy ra v\u1edbi -1 < m < 0 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 4 nghi\u1ec7m.<\/p>\n C\u00e2u 17:<\/b><\/p>\n Ta c\u00f3: a2<\/sup>\u00a0+ b2<\/sup>\u00a0= 7ab<\/p>\n <\/p>\n L\u1ea5y logarit c\u01a1 s\u1ed1 7 hai v\u1ebf ta c\u00f3<\/p>\n <\/p>\n C\u00e2u 18:<\/b><\/p>\n Ta c\u00f3: log1\/4<\/sub>(x2<\/sup>\u00a0– x4<\/sup>) = 1<\/p>\n <\/p>\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m.<\/p>\n C\u00e2u 19:<\/b><\/p>\n \u0110i\u1ec1u ki\u1ec7n: -1 < x < 1. Ta c\u00f3:<\/p>\n <\/p>\n Do \u0111\u00f3 lg(1 – x) > \u221a3<\/p>\n C\u00e2u 20:<\/b><\/p>\n \u0110\u1eb7t t = 3x<\/sup>\u00a0(t > 0). B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho tr\u1edf th\u00e0nh<\/p>\n 3t2<\/sup>\u00a0– 12t + 9 \u2264 0 <=> 1 \u2264 t \u2264 3<\/p>\n Suy ra 1 \u2264 3x<\/sup>\u00a0\u2264 3<\/p>\n V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 0 \u2264 x \u2264 1<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 15:\u00a0Cho ha\u0300m s\u00f4\u0301 \u0110\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 co\u0301 ti\u00ea\u0323m c\u00e2\u0323n ngang la\u0300 A. y = 1\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. y = 1\/2\u00a0\u00a0\u00a0D. y = -5 C\u00e2u 16:\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh |x3| – 12|x| = m (v\u01a1\u0301i -1 < m < 0 ) la\u0300 A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4. C\u00e2u 17:\u00a0Cho hai […]<\/p>\n","protected":false},"author":3,"featured_media":26584,"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 15-C<\/td>\n 16-D<\/td>\n 17-B<\/td>\n 18-B<\/td>\n 19-D<\/td>\n 20-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n