D\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0110\u1ec1 thi th\u1eed THPT qu\u1ed1c gia M\u00f4n To\u00e1n n\u0103m 2016 _ Tr\u01b0\u1eddng THPT Tr\u1ea7n Nh\u00e2n T\u00f4ng . Ch\u00fac c\u00e1c b\u1ea1n h\u1ecdc sinh \u00f4n t\u1eadp th\u1eadt t\u1ed1t \u0111\u1ec3 chu\u1ea9n b\u1ecb cho k\u1ef3 thi quan tr\u1ecdng n\u00e0y!<\/p>\n
K\u1ef2 THI TH\u1eec THPT QU\u1ed0C GIA N\u0102M 2016 \u2013 L\u1ea6N I<\/strong><\/p>\n TR\u01af\u1edcNG THPT TR\u1ea6N NH\u00c2N T\u00d4NG <\/strong><\/p>\n M\u00f4n thi: TO\u00c1N (\u0110\u1ec0 THI CH\u00cdNH TH\u1ee8C)<\/strong><\/p>\n Th\u1eddi gian: 180 ph\u00fat, kh\u00f4ng k\u1ec3 th\u1eddi gian giao \u0111\u1ec1<\/em><\/p>\n C\u00e2u 1 (1 \u0111i\u1ec3m)<\/strong> Kh\u1ea3o s\u00e1t v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup> \u2013 3x2<\/sup> + 4.<\/p>\n C\u00e2u 2 (1 \u0111i\u1ec3m)<\/strong> T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1<\/p>\n <\/a><\/p>\n C\u00e2u 3 (1 \u0111i\u1ec3m) <\/strong>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n <\/a><\/p>\n C\u00e2u 4 (1 \u0111i\u1ec3m) <\/strong>T\u00ednh<\/p>\n <\/a><\/p>\n C\u00e2u 5 (1 \u0111i\u1ec3m)<\/strong><\/p>\n Cho h\u00ecnh ch\u00f3p SABCD c\u00f3 \u0111\u00e1y ABCD l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt.Hai m\u1eb7t ph\u1eb3ng (SAB) v\u00e0 (SAC) c\u00f9ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABCD). Bi\u1ebft r\u1eb1ng AB = a , BC = a\u221a3. v\u00e0 g\u00f3c gi\u1eefa SC v\u1edbi (ABCD) b\u1eb1ng 600<\/sup>. T\u00ednh th\u1ec3 t\u00edch kh\u1ed1i ch\u00f3p SABCD v\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng CE v\u00e0 SB trong \u0111\u00f3 E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a SD.<\/p>\n C\u00e2u 6 (1 \u0111i\u1ec3m)<\/strong><\/p>\n Trong kh\u00f4ng gian cho tam gi\u00e1c ABC c\u00f3 A(1;-1;3); B(-2;3;3); C(1;7;-3) l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (ABC) v\u00e0 t\u00ecm ch\u00e2n \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong k\u1ebb t\u1eeb A tr\u00ean c\u1ea1nh BC.<\/p>\n C\u00e2u 7 (1 \u0111i\u1ec3m)<\/strong><\/p>\n a, M\u1ed9t \u0111o\u00e0n g\u1ed3m 30 ng\u01b0\u1eddi Vi\u1ec7t Nam \u0111i du l\u1ecbch b\u1ecb l\u1ea1c t\u1ea1i Ch\u00e2u Phi, bi\u1ebft r\u1eb3ng trong \u0111o\u00e0n c\u00f3 12 ng\u01b0\u1eddi bi\u1ebft ti\u1ebfng Anh, c\u00f3 8 ng\u01b0\u1eddi bi\u1ebft ti\u1ebfng Ph\u00e1p v\u00e0 c\u00f3 17 ng\u01b0\u1eddi ch\u1ec9 bi\u1ebft ti\u1ebfng Vi\u1ec7t. C\u1ea7n ch\u1ecdn ra 4 ng\u01b0\u1eddi \u0111i h\u1ecfi \u0111\u01b0\u1eddng. T\u00ednh x\u00e1c su\u1ea5t trong 4 ng\u01b0\u1eddi \u0111\u01b0\u1ee3c ch\u1ecdn c\u00f3 2 ng\u01b0\u1eddi bi\u1ebft c\u1ea3 2 th\u1ee9 ti\u1ebfng Anh v\u00e0 Ph\u00e1p.<\/p>\n b, T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c P = (2cos2<\/sup>x \u2013 5)(3 \u2013 2sin2<\/sup>x) bi\u1ebft tanx = 2.<\/p>\n C\u00e2u 8 (1 \u0111i\u1ec3m)<\/strong><\/p>\n Trong m\u1eb7t ph\u1eb3ng to\u1ea1 \u0111\u1ed9 (Oxy), cho h\u00ecnh vu\u00f4ng ABCD.\u0110i\u1ec3m M n\u1eb1m tr\u00ean \u0111o\u1ea1n BC, \u0111\u01b0\u1eddng th\u1eb3ng AM c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x + 3y \u2013 5 = 0, N l\u00e0 \u0111i\u1ec3m tr\u00ean \u0111o\u1ea1n CD sao cho g\u00f3c BMA = AMN. T\u00ecm t\u1ecda \u0111\u1ed9 A bi\u1ebft \u0111\u01b0\u1eddng th\u1eb3ng AN qua \u0111i\u1ec3m K(1;-2).<\/p>\n <\/p>\n <\/a><\/p>\n \u0110\u00c1P \u00c1N \u0110\u1ec0 THI<\/strong><\/p>\n <\/a><\/p>\n <\/a><\/p>\n <\/a><\/p>\n <\/a><\/p>\n <\/a><\/p>\n <\/a><\/p>\n