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{"id":44382,"date":"2019-10-16T23:11:24","date_gmt":"2019-10-16T16:11:24","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=44382"},"modified":"2019-10-16T23:11:38","modified_gmt":"2019-10-16T16:11:38","slug":"giai-toan-10-chuong-1-vecto-bai-4-he-truc-toa-do","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-1-vecto-bai-4-he-truc-toa-do\/","title":{"rendered":"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 1: Vect\u01a1\/ B\u00e0i 4: H\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9"},"content":{"rendered":"\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 H\u00ecnh h\u1ecdc B\u00e0i 4 trang 21<\/strong>: H\u00e3y t\u00ecm c\u00e1ch x\u00e1c \u0111\u1ecbnh v\u1ecb tr\u00ed qu\u00e2n xe v\u00e0 qu\u00e2n m\u00e3 tr\u00ean b\u00e0n c\u1edd vua (h.1.21).<\/ins><\/p>\n\n\n\n

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

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

V\u1ecb tr\u00ed c\u1ee7a qu\u00e2n xe: h\u00e0ng 3, c\u1ed9t c<\/p>\n\n\n\n

V\u1ecb tr\u00ed c\u1ee7a qu\u00e2n m\u00e3: h\u00e0ng 5, c\u1ed9t f<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 H\u00ecnh h\u1ecdc B\u00e0i 4 trang 22<\/strong>: H\u00e3y ph\u00e2n t\u00edch c\u00e1c vect\u01a1 a\u2192<\/em>, b\u2192<\/em> theo hai vect\u01a1 i\u2192<\/em> v\u00e0 j\u2192<\/em> trong h\u00ecnh (h.1.23)<\/ins><\/p>\n\n\n\n

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

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

a\u2192<\/em> = 4i\u2192<\/em> + 2j\u2192<\/em><\/p>\n\n\n\n

b\u2192<\/em> = 0i\u2192<\/em> – 4j<\/em><\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 H\u00ecnh h\u1ecdc B\u00e0i 4 trang 24<\/strong>:\n C\u00e2u h\u1ecfi 3 trang 24: T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a c\u00e1c \u0111i\u1ec3m A, B, C trong h\u00ecnh 1.26. \nCho ba \u0111i\u1ec3m D(-2; 3), E(0; -4), F(3; 0). H\u00e3y v\u1ebd c\u00e1c \u0111i\u1ec3m D, E, F tr\u00ean \nm\u1eb7t ph\u1eb3ng Oxy.<\/ins><\/p>\n\n\n\n

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

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

A(4; 2)<\/p>\n\n\n\n

B(3; 0)<\/p>\n\n\n\n

C(0; 2)<\/ins><\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 H\u00ecnh h\u1ecdc B\u00e0i 4 trang 24<\/strong>: H\u00e3y ch\u1ee9ng minh c\u00f4ng th\u1ee9c tr\u00ean.<\/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 H\u00ecnh h\u1ecdc B\u00e0i 4 trang 25<\/strong>: G\u1ecdi G l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c ABC. H\u00e3y ph\u00e2n t\u00edch vect\u01a1 OG\u2192<\/em> theo ba vect\u01a1 OA\u2192<\/em>, OB\u2192<\/em>, OC\u2192<\/em>. T\u1eeb \u0111\u00f3 h\u00e3y t\u00ednh t\u1ecda \u0111\u1ed9 \u0111i\u1ec3m G theo t\u1ecda \u0111\u1ed9 c\u1ee7a A, B v\u00e0 C.<\/ins><\/p>\n\n\n\n

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

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

B\u00e0i 1 (trang 26 SGK H\u00ecnh h\u1ecdc 10):<\/strong> Tr\u00ean tr\u1ee5c (O, e\u2192) cho c\u00e1c \u0111i\u1ec3m A, B, M, N c\u00f3 t\u1ecda \u0111\u1ed9 l\u1ea7n l\u01b0\u1ee3t l\u00e0 -1, 2, 3, -2<\/p>\n\n\n\n

a) H\u00e3y v\u1ebd tr\u1ee5c v\u00e0 bi\u1ec3u di\u1ec5n c\u00e1c \u0111i\u1ec3m \u0111\u00e3 cho tr\u00ean tr\u1ee5c;<\/ins><\/p>\n\n\n\n

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

b) T\u00ednh \u0111\u1ed9 d\u00e0i \u0111\u1ea1i s\u1ed1 c\u1ee7a \n. T\u1eeb \u0111\u00f3 suy ra hai vec t\u01a1 \n ng\u01b0\u1ee3c h\u01b0\u1edbng. <\/p>\n\n\n\n

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

a) V\u1ebd tr\u1ee5c v\u00e0 bi\u1ec3u di\u1ec5n c\u00e1c \u0111i\u1ec3m <\/p>\n\n\n\n

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

b) Ta c\u00f3:<\/p>\n\n\n\n

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

A c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 \u20131, B c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 2 n\u00ean \n<\/p>\n\n\n\n

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

M c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 3, N c\u00f3 t\u1ecda \u0111\u1ed9 l\u00e0 \u20132 n\u00ean \n<\/p>\n\n\n\n

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

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

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

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

+ N\u1ebfu hai \u0111i\u1ec3m A v\u00e0 B tr\u00ean tr\u1ee5c (O, e\u2192) c\u00f3 t\u1ecda \u0111\u1ed9 l\u1ea7n l\u01b0\u1ee3t l\u00e0 a, b th\u00ec \n<\/p>\n\n\n\n

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

B\u00e0i 2 (trang 26 SGK H\u00ecnh h\u1ecdc 10):<\/strong> Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9, c\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/ins><\/p>\n\n\n\n

a) a\u2192 (-3; 0) v\u00e0 i\u2192 (1; 0) l\u00e0 hai vec t\u01a1 ng\u01b0\u1ee3c h\u01b0\u1edbng. <\/p>\n\n\n\n

b) a\u2192 (3; 4) v\u00e0 b\u2192 (-3; -4) l\u00e0 hai vec t\u01a1 \u0111\u1ed1i nhau <\/p>\n\n\n\n

c) a\u2192 (5; 3) v\u00e0 b\u2192 (3; 5) l\u00e0 hai vec t\u01a1 \u0111\u1ed1i nhau. <\/p>\n\n\n\n

d) Hai vec t\u01a1 b\u1eb1ng nhau khi v\u00e0 ch\u1ec9 khi ch\u00fang c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng nhau v\u00e0 tung \u0111\u1ed9 b\u1eb1ng nhau. <\/p>\n\n\n\n

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

a) \u0110\u00fang<\/p>\n\n\n\n

Gi\u1ea3i th\u00edch: Nh\u1eadn th\u1ea5y a\u2192 = -3.i\u2192<\/p>\n\n\n\n

V\u00ec \u20133 < 0 n\u00ean a\u2192 v\u00e0 i\u2192 ng\u01b0\u1ee3c h\u01b0\u1edbng. <\/p>\n\n\n\n

b) \u0110\u00fang. <\/p>\n\n\n\n

Gi\u1ea3i th\u00edch: <\/p>\n\n\n\n

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

\u21d2 a\u2192 = -b\u2192 n\u00ean a\u2192 v\u00e0 b\u2192 l\u00e0 hai vec t\u01a1 \u0111\u1ed1i nhau. <\/ins><\/p>\n\n\n\n

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

Gi\u1ea3i th\u00edch: <\/p>\n\n\n\n

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

\u21d2 a\u2192 \u2260 -b\u2192 n\u00ean a\u2192 v\u00e0 b\u2192 kh\u00f4ng ph\u1ea3i l\u00e0 hai vec t\u01a1 \u0111\u1ed1i nhau. <\/p>\n\n\n\n

d) \u0110\u00fang<\/p>\n\n\n\n

Nh\u1eadn x\u00e9t SGK : Hai vec t\u01a1 b\u1eb1ng nhau khi v\u00e0 ch\u1ec9 khi ch\u00fang c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng nhau v\u00e0 tung \u0111\u1ed9 b\u1eb1ng nhau. <\/p>\n\n\n\n

B\u00e0i 3 (trang 26 SGK H\u00ecnh h\u1ecdc 10):<\/strong> T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a c\u00e1c vect\u01a1 sau:<\/p>\n\n\n\n

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

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

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

B\u00e0i 4 (trang 26 SGK H\u00ecnh h\u1ecdc 10):<\/strong> Trong m\u1eb7t ph\u1eb3ng Oxy. C\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00fang hay sai?<\/p>\n\n\n\n

a) T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m A b\u1eb1ng t\u1ecda \u0111\u1ed9 c\u1ee7a vect\u01a1 OA;<\/p>\n\n\n\n

b) \u0110i\u1ec3m A n\u1eb1m tr\u00ean tr\u1ee5c ho\u00e0nh th\u00ec c\u00f3 tung \u0111\u1ed9 b\u1eb1ng 0;<\/ins><\/p>\n\n\n\n

c) \u0110i\u1ec3m A n\u1eb1m tr\u00ean tr\u1ee5c tung th\u00ec c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng 0;<\/p>\n\n\n\n

d) Ho\u00e0nh \u0111\u1ed9 v\u00e0 tung \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m A b\u1eb1ng nhau khi v\u00e0 ch\u1ec9 khi A n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c ph\u1ea7n t\u01b0 th\u1ee9 nh\u1ea5t.<\/p>\n\n\n\n

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

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

a) \u0110\u00fang<\/strong>. Gi\u1ea3 s\u1eed A(a; b); O(0; 0) <\/p>\n\n\n\n

b) \u0110\u00fang<\/strong><\/p>\n\n\n\n

c) \u0110\u00fang<\/strong><\/p>\n\n\n\n

d) \u0110\u00fang<\/strong> V\u00ec tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c ph\u1ea7n t\u01b0 th\u1ee9 nh\u1ea5t l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng y = x. <\/p>\n\n\n\n

B\u00e0i 5 (trang 27 SGK H\u00ecnh h\u1ecdc 10):<\/strong> Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy cho \u0111i\u1ec3m M(xo<\/sub>, yo<\/sub>).<\/p>\n\n\n\n

a) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m A \u0111\u1ed1i x\u1ee9ng v\u1edbi M qua tr\u1ee5c Ox;<\/p>\n\n\n\n

b) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m B \u0111\u1ed1i x\u1ee9ng v\u1edbi M qua tr\u1ee5c Oy;<\/ins><\/p>\n\n\n\n

c) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m C \u0111\u1ed1i x\u1ee9ng v\u1edbi M g\u1ed1c O.<\/p>\n\n\n\n

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

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

Bi\u1ec3u di\u1ec5n c\u00e1c \u0111i\u1ec3m tr\u00ean h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 ta th\u1ea5y: <\/p>\n\n\n\n

a) \u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi M(x0<\/sub>; y0<\/sub>) qua tr\u1ee5c Ox l\u00e0 A(x0<\/sub> ; \u2013y0<\/sub>)<\/p>\n\n\n\n

b) \u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi M(x0<\/sub> ; y0<\/sub>) qua tr\u1ee5c Oy l\u00e0 B(\u2013x0<\/sub> ; y0<\/sub>)<\/p>\n\n\n\n

c) \u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi M(x0<\/sub> ; y0<\/sub>) qua g\u1ed1c O l\u00e0 C(\u2013x0<\/sub> ; \u2013y0<\/sub>).<\/p>\n\n\n\n

B\u00e0i 6 (trang 27 SGK H\u00ecnh h\u1ecdc 10):<\/strong> Cho h\u00ecnh b\u00ecnh h\u00e0nh ABCD c\u00f3 A(-1; -2), B(3;2), C(4; -1). T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111\u1ec9nh D.<\/p>\n\n\n\n

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

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

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

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

B\u00e0i 7 (trang 27 SGK H\u00ecnh h\u1ecdc 10):<\/strong> C\u00e1c \n\u0111i\u1ec3m A'(-4; 1), B'(2; 4), C'(2; -2) l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u00e1c c\u1ea1nh BC, \nCA v\u00e0 AB c\u1ee7a tam gi\u00e1c ABC. T\u00ednh t\u1ecda \u0111\u1ed9 c\u00e1c \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c ABC. Ch\u1ee9ng \nminh r\u1eb1ng tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c ABC v\u00e0 A’B’C’ tr\u00f9ng nhau.<\/p>\n\n\n\n

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

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

A\u2019 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC \n<\/p>\n\n\n\n

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

B\u2019 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AC \n<\/p>\n\n\n\n

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

C\u2019 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BA \n<\/p>\n\n\n\n

G\u1ecdi G l\u00e0 tr\u1ecdng t\u00e2m \u0394ABC v\u00e0 G\u2019 l\u00e0 tr\u1ecdng t\u00e2m \u0394A\u2019B\u2019C\u2019<\/ins><\/p>\n\n\n\n

Ta c\u00f3 : <\/p>\n\n\n\n

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

V\u1eady G \u2261 G\u2019 (\u0111pcm)<\/p>\n\n\n\n

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

+ I l\u00e0 trung \u0111i\u1ec3m AB th\u00ec \n<\/p>\n\n\n\n

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

+ G l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c \n<\/p>\n\n\n\n

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

B\u00e0i 8 (trang 27 SGK H\u00ecnh h\u1ecdc 10):<\/strong> Cho vect\u01a1 a = (2; -2), vect\u01a1 b = (1; 4). H\u00e3y ph\u00e2n t\u00edch vect\u01a1 c (5; 0) theo hai vect\u01a1 a v\u00e0 b.<\/p>\n\n\n\n

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

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

<\/ins><\/p>\n","protected":false},"excerpt":{"rendered":"

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 H\u00ecnh h\u1ecdc B\u00e0i 4 trang 21: H\u00e3y t\u00ecm c\u00e1ch x\u00e1c \u0111\u1ecbnh v\u1ecb tr\u00ed qu\u00e2n xe v\u00e0 qu\u00e2n m\u00e3 tr\u00ean b\u00e0n c\u1edd vua (h.1.21). 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