\u0110\u1ec3 h\u1ecdc t\u1ed1t H\u00ecnh h\u1ecdc 12, ph\u1ea7n n\u00e0y gi\u00fap b\u1ea1n gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp trong s\u00e1ch gi\u00e1o khoa To\u00e1n 12 \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n b\u00e1ms\u00e1t theo n\u1ed9i dung s\u00e1ch H\u00ecnh h\u1ecdc 12.<\/strong><\/p>\n B\u00e0i 1:\u00a0Trong kh\u00f4ng gian Oxyz cho 3 vect\u01a1: a = (-1; 1; 0), b =(1; 1; 0) v\u00e0 c = (1; 1; 1). S\u1eed d\u1ee5ng gi\u1ea3 thi\u1ebft n\u00e0y \u0111\u1ec3 tr\u1ea3 l\u1eddi c\u00e1c c\u00e2u h\u1ecfi 1, 2 v\u00e0 3 sau \u0111\u00e2y.<\/strong><\/span><\/p>\n B\u00e0i 1 (trang 94 SGK H\u00ecnh h\u1ecdc 12):<\/b>\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau, m\u1ec7nh \u0111\u1ec1 n\u00e0o sai?<\/p>\n <\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n A.\u0110\u00fang \u00a0\u00a0\u00a0B.\u0110\u00fang\u00a0\u00a0\u00a0C.\u0110\u00fang<\/p>\n \u0110\u00e1p \u00e1n ABC \u0111\u00fang.<\/p>\n <\/p>\n B\u00e0i 2 (trang 94 SGK H\u00ecnh h\u1ecdc 12):\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau, m\u1ec7nh \u0111\u1ec1 n\u00e0o \u0111\u00fang?<\/strong><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n <\/p>\n B\u00e0i 3:\u00a0Cho h\u00ecnh b\u00ecnh h\u00e0nh c\u00f3<\/span><\/strong><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n Ta c\u00f3: A(-1; 1; 0) v\u00e0 B( 1; 1; 0)<\/p>\n G\u1ecdi I l\u00e0 t\u00e2m c\u1ee7a h\u00ecnh b\u00ecnh h\u00e0nh OADB. Khi \u0111\u00f3 I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AB. V\u1eady I(0; 1; 0).<\/p>\n \u00a0B\u00e0i 4 (trang 94 SGK H\u00ecnh h\u1ecdc 12):<\/b>\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau, m\u1ec7nh \u0111\u1ec1 n\u00e0o sai?<\/span><\/p>\n A)B\u1ed1n \u0111i\u1ec3m A, B, C, D t\u1ea1o th\u00e0nh m\u1ed9t t\u1ee9 di\u1ec7n.<\/p>\n B)Tam gi\u00e1c ABD l\u00e0 tam gi\u00e1c \u0111\u1ec1u.<\/p>\n C)AB \u22a5 CD.<\/p>\n D)Tam gi\u00e1c BCD l\u00e0 tam gi\u00e1c vu\u00f4ng.<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n B\u00e0i 5 (trang 95 SGK H\u00ecnh h\u1ecdc 12):<\/b><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n <\/p>\n<\/div>\n B\u00e0i 6 (trang 95 SGK H\u00ecnh h\u1ecdc 12):\u00a0M\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp t\u1ee9 gi\u00e1c ABCD c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng:<\/strong><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n <\/p>\n B\u00e0i 7 (trang 95 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho m\u1eb7t ph\u1eb3ng (\u03b1) \u0111i qua \u0111i\u1ec3m M(0; 0; -1) v\u00e0 song song v\u1edbi gi\u00e1 c\u1ee7a hai vect\u01a1 …<\/strong><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n <\/p>\n B\u00e0i 8 (trang 95 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba \u0111i\u1ec3m A(0; 2; 1), B(3; 0; 1), C(1; 0; 0). Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (ABC) l\u00e0:<\/strong><\/span><\/p>\n A) 2x \u2013 3y \u2013 4z + 2 = 0<\/p>\n B) 2x + 3y \u2013 4z \u2013 2 = 0<\/p>\n C) 4x + 6y \u2013 8z + 2 = 0<\/p>\n D) 2x \u2013 3y \u2013 4z + 1 = 0<\/p>\n L\u1eddi gi\u1ea3i:<\/b><\/p>\n <\/p>\n B\u00e0i 9 (trang 95 SGK H\u00ecnh h\u1ecdc 12):\u00a0G\u1ecdi (\u03b1) l\u00e0 m\u1eb7t ph\u1eb3ng c\u1eaft ba tr\u1ee5c t\u1ecda \u0111\u1ed9 t\u1ea1i ba \u0111i\u1ec3m M(8; 0; 0), N(0; -2; 0), P(0; 0; 4). Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a (\u03b1) l\u00e0:<\/strong><\/span><\/p>\n B\u00e0i 10 (trang 95 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba m\u1eb7t ph\u1eb3ng …<\/strong><\/span><\/p>\n <\/p>\n L\u1eddi gi\u1ea3i<\/strong><\/p>\n <\/p>\n B\u00e0i 11 (trang 96 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m M(2; 0; -1) v\u00e0 c\u00f3 vecto ch\u1ec9 ph\u01b0\u01a1ng a = (4; -6; 2). Ph\u01b0\u01a1ng tr\u00ecnh tham s\u1ed1 c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng \u0394 l\u00e0:<\/strong><\/span><\/p>\n B\u00e0i 12 (trang 96 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho d l\u00e0 d\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m A(1; 2; 3) v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng(\u03b1): 4x + 3y \u2013 7z + 1 =0. Ph\u01b0\u01a1ng tr\u00ecnh tham s\u1ed1 c\u1ee7a d l\u00e0:<\/strong><\/span><\/p>\n B\u00e0i 15 (trang 97 SGK H\u00ecnh h\u1ecdc 12):\u00a0G\u1ecdi S l\u00e0 m\u1eb7t c\u1ea7u t\u00e2m I(2; 1; -1) v\u00e0 ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t ph\u1eb3ng (\u03b1) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 2x – 2y – z + 3 = 0. B\u00e1n k\u00ednh c\u1ee7a (S) b\u1eb1ng:<\/strong><\/span><\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" \u0110\u1ec3 h\u1ecdc t\u1ed1t H\u00ecnh h\u1ecdc 12, ph\u1ea7n n\u00e0y gi\u00fap b\u1ea1n gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp trong s\u00e1ch gi\u00e1o khoa To\u00e1n 12 \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n b\u00e1ms\u00e1t theo n\u1ed9i dung s\u00e1ch H\u00ecnh h\u1ecdc 12. B\u00e0i 1:\u00a0Trong kh\u00f4ng gian Oxyz cho 3 vect\u01a1: a = (-1; 1; 0), b =(1; 1; 0) v\u00e0 c = (1; 1; 1). S\u1eed […]<\/p>\n","protected":false},"author":3,"featured_media":20281,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1298],"tags":[1356,1355,1376],"yoast_head":"\n