C\u00e2u 21:<\/b>\u00a0T\u00e2\u0323p h\u01a1\u0323p nghi\u00ea\u0323m cu\u0309a b\u00e2\u0301t ph\u01b0\u01a1ng tri\u0300nh 3x<\/sup>\u00a0+ 4x<\/sup>\u00a0> 5x<\/sup>\u00a0la\u0300<\/p>\n A. (-\u221e; 2)\u00a0\u00a0\u00a0 B. (0; 2)\u00a0\u00a0\u00a0C. (2; +\u221e)\u00a0\u00a0\u00a0D. (0; 2]<\/p>\n C\u00e2u 22:<\/b>\u00a0T\u00e2\u0323p h\u01a1\u0323p nghi\u00ea\u0323m cu\u0309a b\u00e2\u0301t ph\u01b0\u01a1ng tri\u0300nh<\/p>\n <\/p>\n C\u00e2u 23:<\/b>\u00a0T\u00e2\u0323p h\u01a1\u0323p nghi\u00ea\u0323m cu\u0309a b\u00e2\u0301t ph\u01b0\u01a1ng tri\u0300nh<\/p>\n <\/p>\n A. (-3; 4) \u00a0\u00a0\u00a0B. (-3; 1) \u222a (1; 4) \u00a0\u00a0\u00a0C. (0; 4) \u00a0\u00a0\u00a0D. (0; 1) \u222a (1; 4)<\/p>\n C\u00e2u 24:<\/b>\u00a0Ho\u0323 nguy\u00ean ha\u0300m cu\u0309a ha\u0300m s\u00f4\u0301 y = (2x + 1)5<\/sup>\u00a0la\u0300<\/p>\n <\/p>\n C\u00e2u 25:<\/b>\u00a0Di\u00ea\u0323n ti\u0301ch cu\u0309a hi\u0300nh ph\u0103\u0309ng gi\u01a1\u0301i ha\u0323n b\u01a1\u0309i \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng y = x va\u0300 \u0111\u01b0\u01a1\u0300ng cong y = x2<\/sup>\u00a0b\u0103\u0300ng<\/p>\n <\/p>\n C\u00e2u 26:<\/b>\u00a0Th\u00ea\u0309 ti\u0301ch cu\u0309a v\u00e2\u0323t th\u00ea\u0309 tro\u0300n xoay sinh ra b\u01a1\u0309i phe\u0301p quay quanh tru\u0323c Ox cu\u0309a hi\u0300nh ph\u0103\u0309ng gi\u01a1\u0301i ha\u0323n b\u01a1\u0309i ca\u0301c \u0111\u01b0\u01a1\u0300ng y = 4\/x va\u0300 y = -x + 5 la\u0300<\/p>\n A. 8\u03c0\u00a0\u00a0\u00a0B. 9\u03c0 \u00a0\u00a0\u00a0C. 10\u03c0 \u00a0\u00a0\u00a0 D. 12\u03c0<\/p>\n C\u00e2u 27:<\/b>\u00a0S\u00f4\u0301 na\u0300o sau \u0111\u00e2y la\u0300 s\u00f4\u0301 thu\u00e2\u0300n a\u0309o?<\/p>\n A. (2 + 3i)(2 – 3i)\u00a0\u00a0\u00a0B. (2 + 3i) + (3 – 2i)<\/p>\n C. (2 + 3i) – 2(2 – 3i)<\/p>\n <\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 21:<\/b><\/p>\n B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi<\/p>\n <\/p>\n Do h\u00e0m s\u1ed1<\/p>\n <\/p>\n ngh\u1ecbch bi\u1ebfn tr\u00ean R v\u00e0 f(2) = 1 n\u00ean (1) t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi x < 2<\/p>\n C\u00e2u 22:<\/b><\/p>\n \u0110i\u1ec1u ki\u1ec7n: x > 0, x \u2260 1\/3. Khi \u0111\u00f3 ta c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p:<\/p>\n <\/p>\n V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 (1\/3; 1\/2)<\/p>\n C\u00e2u 23:<\/b><\/p>\n \u0110i\u1ec1u ki\u1ec7n: x > 0, x \u2260 1. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho t\u01b0\u01a1ng \u1ee9ng v\u1edbi<\/p>\n (1\/2)log2<\/sub>(x + 12) > log2<\/sub>x <=> x + 12 > x2<\/sup>\u00a0<=> 0 < x < 4<\/p>\n K\u1ebft h\u1ee3p \u0111i\u1ec1u ki\u1ec7n ta c\u00f3 t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 (0; 1) \u222a (1; 4)<\/p>\n C\u00e2u 24:<\/b><\/p>\n \u0110\u1eb7t t = 2x + 1. Ta c\u00f3: dt = 2dx. Do \u0111\u00f3:<\/p>\n <\/p>\n C\u00e2u 25:<\/b><\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m:<\/p>\n x = x2<\/sup>\u00a0<=> x = 0, x = 1<\/p>\n Di\u1ec7n t\u00edch c\u1ea7n t\u00ednh l\u00e0<\/p>\n <\/p>\n C\u00e2u 26:<\/b><\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m<\/p>\n 4\/x = -x + 5 <=> x2<\/sup>\u00a0– 5x + 4 = 0 <=> x = 1, x = 4<\/p>\n Th\u1ec3 t\u00edch c\u1ea7n t\u00ecm l\u00e0<\/p>\n <\/p>\n C\u00e2u 27:<\/b><\/p>\n Ta c\u00f3:<\/p>\n (2 + 3i)(2 – 3i) = 4 – 9i2<\/sup>\u00a0= 13<\/p>\n (2 + 3i)+ (2 – 3i) = (2 + 2) + (3 – 3)i = 4<\/p>\n (2 + 3i) – (2 – 3i) = 2 – 2 + (3 + 3)i = 6i<\/p>\n <\/p>\n V\u1eady ch\u1ec9 c\u00f3 (2 + 3i) – (2 – 3i) l\u00e0 s\u1ed1 thu\u1ea7n \u1ea3o.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 21:\u00a0T\u00e2\u0323p h\u01a1\u0323p nghi\u00ea\u0323m cu\u0309a b\u00e2\u0301t ph\u01b0\u01a1ng tri\u0300nh 3x\u00a0+ 4x\u00a0> 5x\u00a0la\u0300 A. (-\u221e; 2)\u00a0\u00a0\u00a0 B. (0; 2)\u00a0\u00a0\u00a0C. (2; +\u221e)\u00a0\u00a0\u00a0D. (0; 2] C\u00e2u 22:\u00a0T\u00e2\u0323p h\u01a1\u0323p nghi\u00ea\u0323m cu\u0309a b\u00e2\u0301t ph\u01b0\u01a1ng tri\u0300nh C\u00e2u 23:\u00a0T\u00e2\u0323p h\u01a1\u0323p nghi\u00ea\u0323m cu\u0309a b\u00e2\u0301t ph\u01b0\u01a1ng tri\u0300nh A. (-3; 4) \u00a0\u00a0\u00a0B. (-3; 1) \u222a (1; 4) \u00a0\u00a0\u00a0C. (0; 4) \u00a0\u00a0\u00a0D. (0; 1) \u222a (1; 4) […]<\/p>\n","protected":false},"author":3,"featured_media":26568,"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 21-B<\/td>\n 22-B<\/td>\n 23-D<\/td>\n 24-A<\/td>\n 25-A<\/td>\n 26-B<\/td>\n 27-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n