C\u00e2u 1:<\/b>\u00a0Cho \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 d\u1ea1ng nh\u01b0 h\u00ecnh v\u1ebd.<\/p>\n
<\/p>\n
H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean:<\/p>\n
A. (0;1)<\/p>\n
B. (1;3)<\/p>\n
C. (0; 1) \u222a (1; 3)<\/p>\n
D. (0;1) v\u00e0 (1;3).<\/p>\n
C\u00e2u 2:<\/b>\u00a0H\u1ecfi h\u00e0m s\u1ed1<\/p>\n
<\/p>\n
\u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng n\u00e0o?<\/p>\n
A. (-\u221e ; +\u221e) \u00a0\u00a0\u00a0 B. (-\u221e; -5)<\/p>\n
C. (-5; +\u221e) \u222a (1; 3) \u00a0\u00a0\u00a0D. (0; 1) v\u00e0 (1; 3)<\/p>\n
C\u00e2u 3:<\/b>\u00a0T\u00ecm kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = 2x3<\/sup>\u00a0– 9x2<\/sup>\u00a0+ 12x + 3<\/p>\n A.(-\u221e; 1) \u222a (2; +\u221e) \u00a0\u00a0\u00a0B. (-\u221e 1] v\u00e0 [2; +\u221e)<\/p>\n C. (-\u221e; 1) v\u00e0 (2; +\u221e) \u00a0\u00a0\u00a0 D. (1;2)<\/p>\n C\u00e2u 4:<\/b>\u00a0Kho\u1ea3ng ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 2x2<\/sup>\u00a0– 1 l\u00e0:<\/p>\n A. (-\u221e; -1) v\u00e0 (0; 1) \u00a0\u00a0\u00a0 B. (-\u221e; 0) v\u00e0 (1; +\u221e)<\/p>\n C. (-\u221e; -1) \u222a (0; 1) \u00a0\u00a0\u00a0 D. (0;1)<\/p>\n C\u00e2u 5:<\/b>\u00a0Cho h\u00e0m s\u1ed1<\/p>\n <\/p>\n Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang?<\/p>\n A. H\u00e0m s\u1ed1 (1) ngh\u1ecbch bi\u1ebfn tr\u00ean R\\{1}<\/p>\n B. H\u00e0m s\u1ed1 (1) ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; 1) v\u00e0 (1; +\u221e)<\/p>\n C. H\u00e0m s\u1ed1 (1) ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; 1) \u222a (1; +\u221e)<\/p>\n D. H\u00e0m s\u1ed1 (1) \u0111\u1ed3ng bi\u1ebfn tr\u00ean (-\u221e; 1) v\u00e0 (1; +\u221e)<\/p>\n C\u00e2u 6:<\/b>\u00a0T\u00ecm kho\u1ea3ng \u0111\u1ed3ng bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 f(x)= x + cos2<\/sup>x<\/p>\n A. R\\{0} \u00a0\u00a0\u00a0B. (-\u221e; +\u221e) \u00a0\u00a0\u00a0C. (-1; 1)\u00a0\u00a0\u00a0 D. (0; \u03c0)<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n Tr\u00ean kho\u1ea3ng (0; 1) \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i l\u00ean t\u1eeb tr\u00e1i qua ph\u1ea3i<\/p>\n Tr\u00ean kho\u1ea3ng (1; 3) \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i l\u00ean t\u1eeb tr\u00e1i qua ph\u1ea3i<\/p>\n \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1ecb gi\u00e1n \u0111o\u1ea1n t\u1ea1i x = 1. Do \u0111\u00f3 h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean t\u1eebng kho\u1ea3ng (0; 1) v\u00e0 (1; 3)<\/p>\n C\u00e2u 2:<\/b><\/p>\n H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh \u2200x \u2260 -5<\/p>\n <\/p>\n y’ x\u00e1c \u0111\u1ecbnh \u2200x \u2260 -5 . B\u1ea3ng x\u00e9t d\u1ea5u y\u2019:<\/p>\n <\/p>\n V\u1eady h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; -5) v\u00e0 (-5; +\u221e)<\/p>\n C\u00e2u 3:<\/b><\/p>\n Ta c\u00f3<\/p>\n <\/p>\n B\u1ea3ng x\u00e9t d\u1ea5u \u0111\u1ea1o h\u00e0m:<\/p>\n <\/p>\n H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; 1) v\u00e0 (2; +\u221e)<\/p>\n C\u00e2u 4:<\/b><\/p>\n Ta c\u00f3<\/p>\n <\/p>\n B\u1ea3ng x\u00e9t d\u1ea5u \u0111\u1ea1o h\u00e0m<\/p>\n <\/p>\n H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; -1) v\u00e0 (0; 1)<\/p>\n C\u00e2u 5:<\/b><\/p>\n H\u00e0m s\u1ed1<\/p>\n <\/p>\n x\u00e1c \u0111\u1ecbnh \u2200x \u2260 1<\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n x\u00e1c \u0111\u1ecbnh \u2200x \u2260 1<\/p>\n B\u1ea3ng x\u00e9t d\u1ea5u \u0111\u1ea1o h\u00e0m<\/p>\n <\/p>\n H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e 1) v\u00e0 (1; +\u221e)<\/p>\n C\u00e2u 6:<\/b><\/p>\n f'(x) = 1 – 2sinxcosx = (sinx – cosx)2<\/sup>\u00a0\u2265 0 \u2200x \u2208 R<\/p>\n H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; +\u221e)<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Cho \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 d\u1ea1ng nh\u01b0 h\u00ecnh v\u1ebd. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean: A. (0;1) B. (1;3) C. (0; 1) \u222a (1; 3) D. (0;1) v\u00e0 (1;3). C\u00e2u 2:\u00a0H\u1ecfi h\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng n\u00e0o? A. (-\u221e ; +\u221e) \u00a0\u00a0\u00a0 B. (-\u221e; -5) C. (-5; +\u221e) \u222a (1; 3) \u00a0\u00a0\u00a0D. […]<\/p>\n","protected":false},"author":3,"featured_media":28095,"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 1-D<\/td>\n 2-D<\/td>\n 3-C<\/td>\n 4-A<\/td>\n 5-B<\/td>\n 6-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n