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{"id":44317,"date":"2019-10-15T14:10:22","date_gmt":"2019-10-15T07:10:22","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=44317"},"modified":"2019-10-15T14:10:23","modified_gmt":"2019-10-15T07:10:23","slug":"giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-2-ham-so-y-ax-b","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-2-ham-so-y-ax-b\/","title":{"rendered":"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 2: H\u00e0m s\u1ed1 y = ax + b"},"content":{"rendered":"\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 40<\/strong>: V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1: y = 3x + 2; y = – 1\/2 x+5<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 40<\/strong>: Cho h\u00e0m s\u1ed1 h\u1eb1ng y = 2<\/ins><\/p>\n\n\n\n

X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 t\u1ea1i x = -2; -1; 0; 1; 2.<\/p>\n\n\n\n

Bi\u1ec3u di\u1ec5n c\u00e1c \u0111i\u1ec3m (-2;2), (-1;2), (0;2), (1;2), (2;2) tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9.<\/p>\n\n\n\n

N\u00eau nh\u1eadn x\u00e9t v\u1ec1 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = 2.<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

+) T\u1ea1i x = \u20132; \u20131; 0; 1; 2 th\u00ec y = 2 <\/p>\n\n\n\n

+) \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = 2 l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c ho\u00e0nh v\u00e0 c\u1eaft tr\u1ee5c tung t\u1ea1i \u0111i\u1ec3m (0; 2). <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 1 (trang 41-42 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1:<\/p>\n\n\n\n

a) y = 2x – 3;<\/p>\n\n\n\n

b) y = \u221a2;<\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

d) y = |x| – 1.<\/p>\n\n\n\n

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

a) y = 2x \u2013 3. <\/p>\n\n\n\n

+ x = 1 th\u00ec y = 2.1 \u2013 3 = \u20131. V\u1eady \u0111i\u1ec3m (1 ; \u20131) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. <\/p>\n\n\n\n

+ x = 0 th\u00ec y = 2.0 \u2013 3 = \u20133. V\u1eady \u0111i\u1ec3m (0 ; \u20133) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

b) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = \u221a2 l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c ho\u00e0nh v\u00e0 qua \u0111i\u1ec3m B(0 ; \u221a2)<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

c) <\/ins><\/p>\n\n\n\n

+ x = 2 th\u00ec y = 4. V\u1eady \u0111i\u1ec3m (2; 4) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. <\/p>\n\n\n\n

+ x = 4 th\u00ec y = 1. V\u1eady \u0111i\u1ec3m (4; 1) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

d) y = |x| – 1 hay \n<\/p>\n\n\n\n

V\u1eady \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = |x| \u2013 1 l\u00e0 h\u1ee3p c\u1ee7a hai n\u1eeda: <\/p>\n\n\n\n

+ N\u1eeda \u0111\u1ed3 th\u1ecb l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = x \u2013 1 trong kho\u1ea3ng (0; +\u221e). <\/p>\n\n\n\n

+ N\u1eeda \u0111\u1ed3 th\u1ecb l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = \u2013x \u2013 1 trong kho\u1ea3ng (\u2013\u221e; 0).<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 d\u1ea1ng y = ax + b l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng.<\/p>\n\n\n\n

Do \u0111\u00f3 \u0111\u1ec3 v\u1ebd \u0111\u1ed3 th\u1ecb ta ch\u1ec9 c\u1ea7n x\u00e1c \u0111\u1ecbnh hai \u0111i\u1ec3m thu\u1ed9c \u0111\u1ed3 th\u1ecb r\u1ed3i v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua 2 \u0111i\u1ec3m \u0111\u00f3. <\/p>\n\n\n\n

B\u00e0i 2 (trang 42 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> X\u00e1c \u0111\u1ecbnh a, b \u0111\u1ec3 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax + b \u0111i qua c\u00e1c \u0111i\u1ec3m <\/ins><\/p>\n\n\n\n

a) A(0;3) v\u00e0 B (3\/5; 0)<\/p>\n\n\n\n

b) A(1; 2) v\u00e0 B(2; 1);<\/p>\n\n\n\n

c) A(15; -3) v\u00e0 B(21; -3).<\/p>\n\n\n\n

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

a) A(0;3) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax + b \u21d2 3 = a.0 + b \u21d2 b = 3. <\/p>\n\n\n\n

B (3\/5; 0) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax + b \u21d2 0 = a.3\/5 + 3 \u21d2 a = \u20135. <\/p>\n\n\n\n

V\u1eady a = \u20135; b = 3. <\/p>\n\n\n\n

b) A(1; 2) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax + b \u21d2 2 = a.1 + b \u21d2 b = 2 \u2013 a (1)<\/p>\n\n\n\n

B (2; 1) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax + b \u21d2 1 = 2.a + b (2)<\/p>\n\n\n\n

Thay (1) v\u00e0o (2) ta \u0111\u01b0\u1ee3c: 2a + 2 \u2013 a = 1 \u21d2 a = \u20131 \u21d2 b = 2 \u2013 a = 3. <\/ins><\/p>\n\n\n\n

V\u1eady a = \u20131; b = 3. <\/p>\n\n\n\n

c) A(15; \u20133) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax + b \u21d2 \u20133 = 15.a + b \u21d2 b = \u20133 \u2013 15.a (1)<\/p>\n\n\n\n

B (21; \u20133) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax + b \u21d2 \u20133 = 21.a + b \u21d2 b = \u20133 \u2013 21.a (2)<\/p>\n\n\n\n

T\u1eeb (1) v\u00e0 (2) suy ra \u20133 \u2013 15.a = \u20133 \u2013 21.a \u21d2 a = 0 \u21d2 b = \u20133. <\/p>\n\n\n\n

V\u1eady a = 0; b = \u20133. <\/p>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

\u0110i\u1ec3m A(x0<\/sub>; y0<\/sub>) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = f(x) n\u1ebfu y0<\/sub> = f(x0<\/sub>).<\/p>\n\n\n\n

(Ki\u1ebfn th\u1ee9c l\u1edbp 7). <\/p>\n\n\n\n

B\u00e0i 3 (trang 42 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh y = ax + b c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng:<\/p>\n\n\n\n

a) \u0110i qua hai \u0111i\u1ec3m A(4;3), B(2 ; -1);<\/p>\n\n\n\n

b) \u0110i qua \u0111i\u1ec3m A(1 ; -1) v\u00e0 song song v\u1edbi Ox.<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n

a) <\/p>\n\n\n\n

+ A (4; 3) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b \u21d2 3 = 4.a + b (1)<\/p>\n\n\n\n

+ B (2; \u20131) thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b \u21d2 \u20131 = 2.a + b (2)<\/p>\n\n\n\n

L\u1ea5y (1) tr\u1eeb (2) ta \u0111\u01b0\u1ee3c: 3 \u2013 (\u20131) = (4a + b) \u2013 (2a + b) <\/p>\n\n\n\n

\u21d2 4 = 2a \u21d2 a = 2 \u21d2 b = \u20135. <\/p>\n\n\n\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m A(4;3), B(2 ; \u20131) l\u00e0 y = 2x \u2013 5. <\/p>\n\n\n\n

b) <\/p>\n\n\n\n

+ \u0110\u01b0\u1eddng th\u1eb3ng song song v\u1edbi Ox c\u00f3 d\u1ea1ng y = b.<\/p>\n\n\n\n

+ \u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m A(1 ; \u20131) n\u00ean b = \u2013 1. <\/p>\n\n\n\n

V\u1eady \u0111\u01b0\u1eddng th\u1eb3ng c\u1ea7n t\u00ecm l\u00e0 y = \u20131. <\/p>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

+ \u0110\u01b0\u1eddng th\u1eb3ng y = b (b l\u00e0 h\u1eb1ng s\u1ed1) l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c ho\u00e0nh. <\/p>\n\n\n\n

B\u00e0i 4 (trang 42 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n

a) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0 h\u1ee3p c\u1ee7a hai ph\u1ea7n \u0111\u1ed3 th\u1ecb <\/p>\n\n\n\n

+ Ph\u1ea7n th\u1ee9 nh\u1ea5t l\u00e0 n\u1eeda \u0111\u01b0\u1eddng th\u1eb3ng y = 2x gi\u1eef ph\u1ea7n b\u00ean ph\u1ea3i tr\u1ee5c tung. <\/p>\n\n\n\n

+ Ph\u1ea7n th\u1ee9 hai l\u00e0 n\u1eeda \u0111\u01b0\u1eddng th\u1eb3ng y = \u20131\/2. x gi\u1eef ph\u1ea7n b\u00ean tr\u00e1i tr\u1ee5c tung. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

<\/ins><\/p>\n\n\n\n

b) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0 h\u1ee3p c\u1ee7a hai ph\u1ea7n: <\/p>\n\n\n\n

+ Ph\u1ea7n th\u1ee9 nh\u1ea5t l\u00e0 n\u1eeda \u0111\u01b0\u1eddng th\u1eb3ng x + 1 gi\u1eef l\u1ea1i c\u00e1c \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 \u2265 1.<\/p>\n\n\n\n

+ Ph\u1ea7n th\u1ee9 hai l\u00e0 n\u1eeda \u0111\u01b0\u1eddng th\u1eb3ng \u20132x + 4 gi\u1eef l\u1ea1i c\u00e1c \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 < 1. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n","protected":false},"excerpt":{"rendered":"

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 40: V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1: y = 3x + 2; y = – 1\/2 x+5 L\u1eddi gi\u1ea3i Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 40: Cho h\u00e0m s\u1ed1 h\u1eb1ng y = 2 X\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a […]<\/p>\n","protected":false},"author":12,"featured_media":44318,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 2: H\u00e0m s\u1ed1 y = ax + b<\/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\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-2-ham-so-y-ax-b\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 2: H\u00e0m s\u1ed1 y = ax + b\" \/>\n<meta property=\"og:description\" content=\"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 40: V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1: y = 3x + 2; 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