C\u00e2u 13:<\/b>\u00a0Cho t\u1ee9 di\u1ec7n ABCD c\u00f3 AB = AC = DB = BC = 2a v\u00e0 AD = a . Di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp t\u1ee9 di\u1ec7n l\u00e0:<\/p>\n
<\/p>\n
C\u00e2u 14:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho h\u00ecnh h\u1ed9p ABCD.A’B’C’D’ c\u00f3 A(0;0;1), B(1;2;1), D(2;-1;1), A\u2019(5;2;4). T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m C\u2019 l\u00e0:<\/p>\n
A. (3;1;1)\u00a0\u00a0\u00a0B. (8;3;4)\u00a0\u00a0\u00a0C. (-8;-3;-2)\u00a0\u00a0\u00a0D. (-2;-1;-2)<\/p>\n
C\u00e2u 15:<\/b>\u00a0Cho hai vect\u01a1\u00a0a\u2192<\/i>,\u00a0b\u2192<\/i>\u00a0t\u1ea1o v\u1edbi nhau m\u1ed9t g\u00f3c 60o<\/sup>. Bi\u1ebft \u0111\u1ed9 d\u00e0i c\u1ee7a hai vect\u01a1 \u0111\u00f3 \u0111\u1ec1u l\u00e0 6. \u0110\u1ed9 d\u00e0i c\u1ee7a vect\u01a1 t\u1ed5ng\u00a0a\u2192<\/i>\u00a0+\u00a0b\u2192<\/i>\u00a0l\u00e0:<\/p>\n A. 6\u00a0\u00a0\u00a0B. 12\u00a0\u00a0\u00a0C. 6\u221a2\u00a0\u00a0\u00a0D. 2\u221a27<\/p>\n C\u00e2u 16:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho m\u1eb7t c\u1ea7u (S) c\u00f3 t\u00e2m I(-1;0;1), b\u00e1n k\u00ednh R=5. M\u1eb7t ph\u1eb3ng (P): 4x – 3y + z + m = 9 c\u1eaft m\u1eb7t c\u1ea7u (S) theo m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng 5. Khi \u0111\u00f3 m b\u1eb1ng:<\/p>\n A. m=-1\u00a0\u00a0\u00a0B. m=7\u00a0\u00a0\u00a0C. m=3\u00a0\u00a0\u00a0D. m=9<\/p>\n C\u00e2u 17:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, cho \u0111i\u1ec3m A(2;1;0) v\u00e0 m\u1eb7t ph\u1eb3ng (P) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x + 2y – 2z + m = 0 . Trong c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a m sau \u0111\u00e2y, gi\u00e1 tr\u1ecb n\u00e0o th\u1ecfa m\u00e3n kho\u1ea3ng c\u00e1ch t\u1eeb A \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P) b\u1eb1ng 1?<\/p>\n A. m=6\u00a0\u00a0\u00a0B. m=-1\u00a0\u00a0\u00a0C. m=3\u00a0\u00a0\u00a0D. \u0110\u00e1p s\u1ed1 kh\u00e1c<\/p>\n C\u00e2u 18:<\/b>\u00a0Trong kh\u00f4ng gian Oxyz, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1eb7t ph\u1eb3ng (P) \u0111i qua ba \u0111i\u1ec3m A(-1;0;-1), B(0;1;3), C(-2;-1;-3)<\/p>\n A. x – y + 1 = 0\u00a0\u00a0\u00a0C. x + z + 2 = 0<\/p>\n B. x – y – 1 = 0\u00a0\u00a0\u00a0D. x + y + 1 = 0<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 13:\u00a0Cho t\u1ee9 di\u1ec7n ABCD c\u00f3 AB = AC = DB = BC = 2a v\u00e0 AD = a . Di\u1ec7n t\u00edch c\u1ee7a m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp t\u1ee9 di\u1ec7n l\u00e0: C\u00e2u 14:\u00a0Trong kh\u00f4ng gian Oxyz, cho h\u00ecnh h\u1ed9p ABCD.A’B’C’D’ c\u00f3 A(0;0;1), B(1;2;1), D(2;-1;1), A\u2019(5;2;4). T\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m C\u2019 l\u00e0: A. (3;1;1)\u00a0\u00a0\u00a0B. (8;3;4)\u00a0\u00a0\u00a0C. (-8;-3;-2)\u00a0\u00a0\u00a0D. (-2;-1;-2) […]<\/p>\n","protected":false},"author":3,"featured_media":26765,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1450,1448],"yoast_head":"\n\n\n
\n 13-C<\/td>\n 14-B<\/td>\n 15-D<\/td>\n 16-C<\/td>\n 17-C<\/td>\n 18-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"