C\u00e2u 33:<\/b>\u00a0\u00d4ng A g\u1eedi ti\u1ebft ki\u1ec7m v\u00e0o ng\u00e2n h\u00e0ng 200 tri\u1ec7u \u0111\u1ed3ng v\u1edbi h\u00ecnh th\u1ee9c l\u00e3i k\u00e9p. Sau 5 n\u0103m \u00f4ng r\u00fat h\u1ebft ti\u1ec1n ra \u0111\u01b0\u1ee3c m\u1ed9t kho\u1ea3n 283142000 \u0111\u1ed3ng. H\u1ecfi \u00f4ng A g\u1eedi v\u1edbi l\u00e3i su\u1ea5t bao nhi\u00eau, bi\u1ebft r\u1eb1ng trong th\u1eddi gian \u0111\u00f3 l\u00e3i su\u1ea5t kh\u00f4ng thay \u0111\u1ed5i?<\/p>\n
A. 6,8% m\u1ed9t n\u0103m\u00a0\u00a0\u00a0C. 7,2% m\u1ed9t n\u0103m<\/p>\n
B. 7% m\u1ed9t n\u0103m\u00a0\u00a0\u00a0D. 8% m\u1ed9t n\u0103m<\/p>\n
C\u00e2u 34:<\/b>\u00a0S\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 c\u1ee7a m\u1ed9t m\u1ebb c\u1ea5y vi khu\u1ea9n sau t ng\u00e0y k\u1ec3 t\u1eeb l\u00fac ban \u0111\u1ea7u \u0111\u01b0\u1ee3c \u01b0\u1edbc l\u01b0\u1ee3ng b\u1edfi c\u00f4ng th\u1ee9c N(t) = 1200.(1,48)t<\/sup>\u00a0. Sau bao l\u00e2u th\u00ec s\u1ed1 l\u01b0\u1ee3ng vi khu\u1ea9n \u0111\u1ea1t \u0111\u1ebfn 5000 c\u00e1 th\u1ec3? L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ph\u1ea7n m\u01b0\u1eddi<\/p>\n A. 10,3 ng\u00e0y\u00a0\u00a0\u00a0B. 12,3 ng\u00e0y\u00a0\u00a0\u00a0C. 13,0 ng\u00e0y\u00a0\u00a0\u00a0D. 61,7 ng\u00e0y<\/p>\n C\u00e2u 35:<\/b>\u00a0T\u00ecm t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n <\/p>\n A. (0; 4)\u00a0\u00a0\u00a0C. (-\u221e; 1) \u222a (\u221a2; 4)<\/p>\n B. (\u221a2; 4)\u00a0\u00a0\u00a0D. (0; 1) \u222a (\u221a2; 4)<\/p>\n C\u00e2u 36:<\/b>\u00a0Trong c\u00e1c s\u1ed1 \u0111\u01b0\u1ee3c li\u1ec7t k\u00ea trong b\u1ed1n \u0111\u00e1p \u00e1n A, B, C, D d\u01b0\u1edbi \u0111\u00e2y, s\u1ed1 n\u00e0o b\u00e9 nh\u1ea5t?<\/p>\n <\/p>\n C\u00e2u 37:<\/b>\u00a0T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: P = log(tan1o<\/sup>) + log(tan2o<\/sup>) + log(tan3o<\/sup>) +…+ log(tan88o<\/sup>) + log(tan89o<\/sup>)<\/p>\n <\/p>\n C\u00e2u 38:<\/b>\u00a0Cho p v\u00e0 q l\u00e0 c\u00e1c s\u1ed1 d\u01b0\u01a1ng th\u1ecfa m\u00e3n log9<\/sub>p = log12<\/sub>q = log16<\/sub>(p + q) . T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a q\/p<\/p>\n <\/p>\n C\u00e2u 39:<\/b>\u00a0G\u1ecdi P v\u00e0 Q l\u00e0 hai \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ex\/2<\/sup>\u00a0l\u1ea7n l\u01b0\u1ee3t c\u00f3 ho\u00e0nh \u0111\u1ed9 ln4 v\u00e0 ln16 . K\u00ed hi\u1ec7u l l\u00e0 \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng PQ. H\u1ec7 th\u1ee9c n\u00e0o sau \u0111\u00e2y \u0111\u00fang?<\/p>\n A. l2<\/sup>\u00a0= 4(ln4 + 1)\u00a0\u00a0\u00a0C. l2<\/sup>\u00a0= 4(ln16 + 1)<\/p>\n B. l2<\/sup>\u00a0= 4((ln4)2<\/sup>\u00a0+ 1)\u00a0\u00a0\u00a0D. l2<\/sup>\u00a0= 4((ln2)2<\/sup>\u00a0+ 1)<\/p>\n C\u00e2u 40:<\/b>\u00a0Bi\u1ebft r\u1eb1ng log2<\/sub>(log3<\/sub>(log4<\/sub>x)) = log3<\/sub>(log4<\/sub>(log2<\/sub>y)) = log4<\/sub>(log2<\/sub>(log3<\/sub>z)) = 0. T\u00ednh t\u1ed5ng x + y + z<\/p>\n A. 50\u00a0\u00a0\u00a0B. 58\u00a0\u00a0\u00a0C. 89\u00a0\u00a0\u00a0D. 111<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 33:<\/b><\/p>\n Gi\u1ea3 s\u1eed l\u00e3i su\u1ea5t l\u00e0 r. Ta c\u00f3 200000000.(1 + r)5<\/sup>\u00a0= 283142000<\/p>\n <\/p>\n C\u00e2u 34:<\/b><\/p>\n S\u1ed1 l\u01b0\u1ee3ng vi khu\u1ea9n \u0111\u1ea1t \u0111\u1ebfn 5000 c\u00e1 th\u1ec3 khi 5000 = 1200.(1,148)t<\/sup><\/p>\n <\/p>\n C\u00e2u 35:<\/b><\/p>\n \u0110\u1eb7t t = log2<\/sub>x , nh\u1eadn \u0111\u01b0\u1ee3c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n <\/p>\n C\u00e2u 36:<\/b><\/p>\n Vi\u1ebft c\u00e1c s\u1ed1 h\u1ea1ng v\u1ec1 c\u00f9ng d\u1ea1ng c\u0103n b\u1eadc 300 c\u1ee7a m\u1ed9t bi\u1ec3u th\u1ee9c :<\/p>\n <\/p>\n C\u00e2u 37:<\/b><\/p>\n P = log(tan1o<\/sup>.tan2o<\/sup>.tan3o<\/sup>…tan88o<\/sup>.tan89o<\/sup>\u00a0)<\/p>\n = log((tan1o<\/sup>.tan89o<\/sup>).(tan2o<\/sup>.tan88o<\/sup>)…tan45o<\/sup>)<\/p>\n = log(1.1…1) = log1 = 0<\/p>\n C\u00e2u 38:<\/b><\/p>\n \u0110\u1eb7t log9<\/sub>p = log12<\/sub>q = log16<\/sub>(p + q) = t<\/p>\n => p = 9t<\/sup>, q = 12t<\/sup>, p + q = 16t<\/sup><\/p>\n => 9t<\/sup>\u00a0+ 12t<\/sup>\u00a0= 16t<\/sup>\u00a0hay 32t<\/sup>\u00a0+ 3t<\/sup>.4t<\/sup>\u00a0= 42t<\/sup><\/p>\n Chia c\u1ea3 hai v\u1ebf \u0111\u1eb3ng th\u1ee9c n\u00e0y cho 32t<\/sup>\u00a0ta \u0111\u01b0\u1ee3c<\/p>\n <\/p>\n ta \u0111\u01b0\u1ee3c: X2<\/sup>\u00a0– X – 1 = 0<\/p>\n <\/p>\n C\u00e2u 39:<\/b><\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n Do \u0111\u00f3 P(ln4; 2) v\u00e0 Q(ln16; 4)<\/p>\n T\u1eeb \u0111\u00f3 l2<\/sup>\u00a0= (ln16 – ln4)2<\/sup>\u00a0+ (4 – 2)2<\/sup>\u00a0= (ln4)2<\/sup>\u00a0+ 4 = (2ln2)2<\/sup>\u00a0+ 4 = 4((ln2)2<\/sup>\u00a0+ 1)<\/p>\n C\u00e2u 40:<\/b><\/p>\n T\u1eeb gi\u1ea3 thi\u1ebft t\u00ednh \u0111\u01b0\u1ee3c: x = 43<\/sup>, y = 24<\/sup>, z = 32<\/sup><\/p>\n T\u1eeb \u0111\u00f3 x + y + z = 64 + 16 + 9 = 89<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 33:\u00a0\u00d4ng A g\u1eedi ti\u1ebft ki\u1ec7m v\u00e0o ng\u00e2n h\u00e0ng 200 tri\u1ec7u \u0111\u1ed3ng v\u1edbi h\u00ecnh th\u1ee9c l\u00e3i k\u00e9p. Sau 5 n\u0103m \u00f4ng r\u00fat h\u1ebft ti\u1ec1n ra \u0111\u01b0\u1ee3c m\u1ed9t kho\u1ea3n 283142000 \u0111\u1ed3ng. H\u1ecfi \u00f4ng A g\u1eedi v\u1edbi l\u00e3i su\u1ea5t bao nhi\u00eau, bi\u1ebft r\u1eb1ng trong th\u1eddi gian \u0111\u00f3 l\u00e3i su\u1ea5t kh\u00f4ng thay \u0111\u1ed5i? A. 6,8% m\u1ed9t n\u0103m\u00a0\u00a0\u00a0C. 7,2% […]<\/p>\n","protected":false},"author":3,"featured_media":27242,"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 33-C<\/td>\n 34-A<\/td>\n 35-D<\/td>\n 36-B<\/td>\n 37-B<\/td>\n 38-D<\/td>\n 39-D<\/td>\n 40-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n