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{"id":23918,"date":"2018-03-27T11:38:02","date_gmt":"2018-03-27T04:38:02","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=23918"},"modified":"2018-03-27T11:38:02","modified_gmt":"2018-03-27T04:38:02","slug":"dai-so-chuong-1-bai-5-duong-tiem-can-cua-do-thi-ham-so","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/dai-so-chuong-1-bai-5-duong-tiem-can-cua-do-thi-ham-so\/","title":{"rendered":"\u0110\u1ea1i s\u1ed1 – Ch\u01b0\u01a1ng 1 – B\u00e0i 5: \u0110\u01b0\u1eddng ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1"},"content":{"rendered":"

B\u00e0i 34 (trang 35 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

T\u00ecm c\u00e1c ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 sau:<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

TX\u0110: R \\ {-2\/3}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=1\/3 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->+\u221e v\u00e0 khi x->-\u221e)<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-2\/3 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb x->(-2\/3)–<\/sup><\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-2\/3 l\u00e0 ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->(-2\/3)+<\/sup>)<\/p>\n

b) TX\u0110: R \\ {-3}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=-2 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->+\u221e v\u00e0 khi x->-\u221e).<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-3 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb x \u2192 (-3)–<\/sup><\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-3 l\u00e0 ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x \u2192 (-3)+<\/sup>)<\/p>\n

TX\u0110: R\\ {3}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=3 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->3–<\/sup>\u00a0v\u00e0 khi x->3+<\/sup>\u00a0).<\/p>\n

\u0110\u01b0\u1eddng th\u1eb3ng y=x+2 l\u00e0m ti\u1ec7n c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e) v\u00e0 khi x->+\u221e)<\/p>\n

d) C\u00e1ch 1. H\u01b0\u1edbng d\u1eabn:<\/p>\n

TX\u0110: R \\{-1\/2}<\/p>\n

L\u00e0m t\u01b0\u01a1ng t\u1ef1 c\u00e2u c) \u0111\u1ec3 c\u00f3 y=x\/2 -7\/4 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean, x=-1\/2 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>\n

TX\u0110: R \\ {-1\/2 }<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-1\/2 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x \u2192 (-1\/2)–<\/sup>\u00a0v\u00e0 khi x \u2192 (-1\/2)+<\/sup>).<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=1\/2 x-7\/4 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e)<\/p>\n

e) H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R \\ {\u00b1 1}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=1 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->1–<\/sup>\u00a0v\u00e0 khi x->1+<\/sup>).<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-1 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->(-1)–<\/sup>\u00a0v\u00e0 khi x->(-1)+<\/sup>).<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=0 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e v\u00e0 khi x->+\u221e).<\/p>\n

K\u1ebft lu\u1eadn: \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho c\u00f3 hai ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng x=\u00b11 v\u00e0 m\u1ed9t ti\u1ec7m c\u1eadn ngang l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = 0\u2019<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=0 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e v\u00e0 khi x->+\u221e).<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-1 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->(-1)–<\/sup>\u00a0v\u00e0 khi x->(-1)+<\/sup>).<\/p>\n

K\u1ebft lu\u1eadn: \u0111\u1ed3 th\u1ecb c\u00f3 ti\u1ec7m c\u1eadn ngang l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y =0 v\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng x = -1<\/p>\n

B\u00e0i 35 (trang 35 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

T\u00ecm c\u00e1c ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 sau:<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R \\ {0}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x = 0 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->0+<\/sup>\u00a0v\u00e0 khi x->0–<\/sup>).<\/p>\n

N\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=x-3 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e v\u00e0 khi x->+\u221e)<\/p>\n

H\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean R \\ {0; 2}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x = 0 l\u00e0m ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->0–<\/sup>\u00a0v\u00e0 khi x->0+<\/sup>)<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x = 2 c\u0169ng l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x \u2192 2–<\/sup>\u00a0v\u00e0 khi x \u2192 2+<\/sup>)<\/p>\n

N\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=x+2 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x-> +\u221e)<\/p>\n

T\u01b0\u01a1ng t\u1ef1, y = x + 2 c\u0169ng l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e)<\/p>\n

K\u1ebft lu\u1eadn: \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho c\u00f3 c\u00e1c \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 x = 0; x = 2 v\u00e0 ti\u1ec7m c\u1eadn xi\u00ean l\u00e0 y = x + 2.<\/p>\n

c)TX\u0110: R \\ {\u00b11}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x = -1 c\u0169ng l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->(-1)–<\/sup>\u00a0v\u00e0 khi x->(-1)+<\/sup>)<\/p>\n

T\u01b0\u01a1ng t\u1ef1, d\u01b0\u1edbng th\u1eb3ng x = 1 c\u0169ng l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (khi x->1–<\/sup>\u00a0v\u00e0 khi x->1+<\/sup>)<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y = x l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e v\u00e0 x->+\u221e)<\/p>\n

K\u1ebft lu\u1eadn: \u0111\u1ed3 th\u1ecb c\u00f3 hai ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng x= -1, x = 1 v\u00e0 ti\u1ec7m c\u1eadn xi\u00ean l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = x.<\/p>\n

d) TX\u0110: R \\ {-1;3\/5}<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x = – 1 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->(-1)–<\/sup>\u00a0v\u00e0 khi x->(-1)<\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x = 3\/5 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->(3\/5)–<\/sup>\u00a0v\u00e0 khi x->(3\/5)+<\/sup><\/p>\n

n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y = -1\/5 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e v\u00e0 x->+\u221e)<\/p>\n

K\u1ebft lu\u1eadn: \u0110\u1ed3 th\u1ecb c\u00f3 hai ti\u1ec7m c\u1eadn \u0111\u1ee9ng, l\u00e0 c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng x = -1, x=3\/5 v\u00e0 c\u00f3 ti\u1ec7m c\u1eadn ngang l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y=-1\/5<\/p>\n

B\u00e0i 36 (trang 36 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

T\u00ecm c\u00e1c ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 sau.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

TX\u0110: (-\u221e,-1] \u222a[1; +\u221e)<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = -x l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e)<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = x l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->+\u221e)<\/p>\n

K\u1ebft lu\u1eadn: \u0110\u1ed3 th\u1ecb c\u00f3 ti\u1ec7m c\u1eadn xi\u00ean l\u00e0 y = -x (khi x->-\u221e) v\u00e0 y = x (khi x->+\u221e)<\/p>\n

TX\u0110: (-\u221e,-1] \u222a[1; +\u221e)<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = x l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e)<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = 3x l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->+\u221e)<\/p>\n

K\u1ebft lu\u1eadn: c\u00e1c \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb l\u00e0: y = 0 (khi x->-\u221e),y=3x (khi x->+\u221e)<\/p>\n

c) TX\u0110: R<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = 0 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e)<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = 2x l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->+\u221e)<\/p>\n

K\u1ebft lu\u1eadn: c\u00e1c \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb l\u00e0: y = 0 (khi x->-\u221e),y=2x (khi x->+\u221e)<\/p>\n

d) TX\u0110: R<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = -x – 1\/2 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->-\u221e)<\/p>\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng y = x+1\/2 l\u00e0 ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->+\u221e)<\/p>\n

K\u1ebft lu\u1eadn: c\u00e1c \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xi\u00ean c\u1ee7a \u0111\u1ed3 th\u1ecb l\u00e0: y = -x-1\/2 (khi x->-\u221e),y=x+1\/2 (khi x->+\u221e)<\/p>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 34 (trang 35 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):\u00a0 T\u00ecm c\u00e1c ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 sau: L\u1eddi gi\u1ea3i: TX\u0110: R \\ {-2\/3} n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng y=1\/3 l\u00e0 ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb (khi x->+\u221e v\u00e0 khi x->-\u221e) n\u00ean \u0111\u01b0\u1eddng th\u1eb3ng x=-2\/3 l\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb x->(-2\/3)– n\u00ean […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1302],"tags":[1392,1393],"yoast_head":"\n\u0110\u1ea1i s\u1ed1 - Ch\u01b0\u01a1ng 1 - B\u00e0i 5: \u0110\u01b0\u1eddng ti\u1ec7m c\u1eadn c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/dai-so-chuong-1-bai-5-duong-tiem-can-cua-do-thi-ham-so\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u0110\u1ea1i s\u1ed1 - 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