C\u00e2u 1:<\/b>\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau m\u1ec7nh \u0111\u1ec1 n\u00e0o nh\u1eadn gi\u00e1 tr\u1ecb \u0111\u00fang?<\/p>\n
A. H\u00e0m s\u1ed1 y = 1\/x c\u00f3 nguy\u00ean h\u00e0m tr\u00ean (-\u221e; +\u221e).<\/p>\n
B. 3x2<\/sup>\u00a0l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean h\u00e0m c\u1ee7a x3<\/sup>\u00a0tr\u00ean (-\u221e; +\u221e).<\/p>\n C. H\u00e0m s\u1ed1 y = |x| c\u00f3 nguy\u00ean h\u00e0m tr\u00ean (-\u221e;+\u221e).<\/p>\n D. 1\/x + C l\u00e0 h\u1ecd nguy\u00ean h\u00e0m c\u1ee7a ln\u2061x tr\u00ean (0;+\u221e).<\/p>\n C\u00e2u 2:<\/b>\u00a0H\u00e0m s\u1ed1 n\u00e0o d\u01b0\u1edbi \u0111\u00e2y kh\u00f4ng ph\u1ea3i l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x)=2x-sin\u20612x ?<\/p>\n x2<\/sup>\u00a0+ (1\/2).cos\u20612x \u00a0\u00a0\u00a0 B. x2<\/sup>\u00a0+ cos2<\/sup>\u00a0x \u00a0\u00a0\u00a0 C. x2<\/sup>\u00a0– sin2<\/sup>x \u00a0\u00a0\u00a0 D. x2<\/sup>\u00a0+ cos\u20612x .<\/p>\n C\u00e2u 3:<\/b>\u00a0T\u00ecm nguy\u00ean h\u00e0m c\u1ee7a<\/p>\n <\/p>\n <\/p>\n C\u00e2u 4:<\/b><\/p>\n <\/p>\n <\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00ed: M\u1ecdi h\u00e0m s\u1ed1 li\u00ean t\u1ee5c tr\u00ean K \u0111\u1ec1u c\u00f3 nguy\u00ean<\/p>\n h\u00e0m tr\u00ean K. V\u00ec y = |x| li\u00ean t\u1ee5c tr\u00ean R n\u00ean c\u00f3 nguy\u00ean h\u00e0m tr\u00ean R .<\/p>\n Ph\u01b0\u01a1ng \u00e1n A sai v\u00ec y=1\/x kh\u00f4ng x\u00e1c \u0111\u1ecbnh t\u1ea1i x=0 \u2208 (-\u221e;+\u221e).<\/p>\n Ph\u01b0\u01a1ng \u00e1n B sai v\u00ec 3x2<\/sup>\u00a0l\u00e0 \u0111\u1ea1o h\u00e0m c\u1ee7a x3<\/sup>.<\/p>\n Ph\u01b0\u01a1ng \u00e1n D sai v\u00ec 1\/x l\u00e0 \u0111\u1ea1o h\u00e0m c\u1ee7a ln\u2061x tr\u00ean (0; +\u221e).<\/p>\n V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n C\u00e2u 2:<\/b><\/p>\n Ta c\u00f3<\/p>\n \u222b(2x-sin\u20612x)dx=2\u222bxdx-\u222bsin\u20612xdx<\/p>\n <\/p>\n D kh\u00f4ng ph\u1ea3i l\u00e0 nguy\u00ean h\u00e0m c\u1ee7a f(x). V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D.<\/p>\n C\u00e2u 3:<\/b><\/p>\n V\u1edbi x \u2208 (0; +\u221e) ta c\u00f3<\/p>\n <\/p>\n V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n C\u00e2u 4:<\/b><\/p>\n <\/p>\n V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n B.<\/p>\n Ghi ch\u00fa. Y\u00eau c\u1ea7u t\u00ecm nguy\u00ean h\u00e0m c\u1ee7a m\u1ed9t h\u00e0m s\u1ed1 \u0111\u01b0\u1ee3c hi\u1ec3u l\u00e0 t\u00ecm nguy\u00ean h\u00e0m tr\u00ean t\u1eebng kho\u1ea3ng x\u00e1c \u0111\u1ecbnh c\u1ee7a n\u00f3.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau m\u1ec7nh \u0111\u1ec1 n\u00e0o nh\u1eadn gi\u00e1 tr\u1ecb \u0111\u00fang? A. H\u00e0m s\u1ed1 y = 1\/x c\u00f3 nguy\u00ean h\u00e0m tr\u00ean (-\u221e; +\u221e). B. 3×2\u00a0l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean h\u00e0m c\u1ee7a x3\u00a0tr\u00ean (-\u221e; +\u221e). C. H\u00e0m s\u1ed1 y = |x| c\u00f3 nguy\u00ean h\u00e0m tr\u00ean (-\u221e;+\u221e). D. 1\/x + C l\u00e0 h\u1ecd nguy\u00ean h\u00e0m c\u1ee7a […]<\/p>\n","protected":false},"author":3,"featured_media":27140,"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-C<\/td>\n 2-D<\/td>\n 3-C<\/td>\n 4-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n