D\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0110\u1ec1 thi th\u1eed THPT qu\u1ed1c gia M\u00f4n To\u00e1n n\u0103m 2016 l\u1ea7n 2 _Chuy\u00ean Nguy\u1ec5n Hu\u1ec7. 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
\u0110\u1ec0 THI TH\u1eec L\u1ea6N II\u2013 K\u1ef2 THI THPT QU\u1ed0C GIA<\/strong><\/p>\n N\u0102M H\u1eccC 2015 \u2013 2016 \u2013 M\u00d4N TO\u00c1N<\/strong><\/p>\n Th\u1eddi gian 180 ph\u00fat<\/em><\/p>\n C\u00e2u 1: a) (1 \u0111i\u1ec3m) Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb (C) c\u1ee7a h\u00e0m s\u1ed1: <\/a><\/p>\n C\u00e2u 2 (1,0 \u0111i\u1ec3m).<\/strong><\/p>\n a). Cho h\u00e0m s\u1ed1:<\/p>\n <\/a><\/p>\n b) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau:<\/p>\n <\/a><\/p>\n C\u00e2u 3 (1,0 \u0111i\u1ec3m).<\/strong> T\u00ednh t\u00edch ph\u00e2n:<\/p>\n <\/a><\/p>\n C\u00e2u 4(1,0 \u0111i\u1ec3m)<\/strong>. Trong kh\u00f4ng gian Oxyz<\/em>, cho mp(P<\/em>) v\u00e0 m\u1eb7t c\u1ea7u (S<\/em>) l\u1ea7n l\u01b0\u1ee3t c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n <\/a><\/p>\n Ch\u1ee9ng minh m\u1eb7t c\u1ea7u (S<\/em>) c\u1eaft m\u1eb7t ph\u1eb3ng (P<\/em>) .T\u00ecm t\u1ecda \u0111\u1ed9 t\u00e2m v\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n giao tuy\u1ebfn c\u1ee7a m\u1eb7t c\u1ea7u v\u00e0 m\u1eb7t ph\u1eb3ng.<\/p>\n . C\u00e2u 5(1,0 \u0111i\u1ec3m). <\/strong><\/p>\n <\/a><\/p>\n b)Cho \u0111a gi\u00e1c \u0111\u1ec1u 20 \u0111\u1ec9nh n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O. Ch\u1ecdn ng\u1eabu nhi\u00ean 4 \u0111\u1ec9nh c\u1ee7a \u0111a gi\u00e1c \u0111\u00f3.T\u00ednh x\u00e1c su\u1ea5t sao cho 4 \u0111\u1ec9nh \u0111\u01b0\u1ee3c ch\u1ecdn l\u00e0 4 \u0111\u1ec9nh c\u1ee7a m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt.<\/p>\n C\u00e2u 6(1,0 \u0111i\u1ec3m)<\/strong>. Cho h\u00ecnh ch\u00f3p S<\/em>.ABC<\/em> c\u00f3 m\u1eb7t ph\u1eb3ng (SAC<\/em>) vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABC<\/em>), SA=<\/em> AB <\/em>=a, AC =<\/em> 2a<\/em> v\u00e0 g\u00f3c ASC <\/em>=<\/span>ABC =<\/em>90 .0<\/sup> T\u00ednh th\u1ec3 t\u00edch kh\u1ed1i ch\u00f3p S<\/em>.ABC<\/em> v\u00e0 cosin c\u1ee7a g\u00f3c gi\u1eefa hai m\u1eb7t ph\u1eb3ng (SAB<\/em>), (SBC<\/em>).<\/p>\n C\u00e2u 7 (1,0 \u0111i\u1ec3m)<\/strong>. Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxy, cho h\u00ecnh b\u00ecnh h\u00e0nh ABCD c\u00f3 g\u00f3c BAD<\/em> =1350<\/sup>, tr\u1ef1c t\u00e2m tam gi\u00e1c ABD l\u00e0 H(-1;0).\u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua D v\u00e0 H c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x-<\/em>3y +<\/em>1= 0 . T\u00ecm t\u1ecda \u0111\u1ed9 c\u00e1c \u0111\u1ec9nh c\u1ee7a h\u00ecnh b\u00ecnh h\u00e0nh bi\u1ebft \u0111i\u1ec3m G(5\/3 ;2 ) l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c ADC.<\/p>\n <\/a><\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n \u0110\u00c1P \u00c1N \u0110\u1ec0 THI<\/strong><\/p>\n <\/a><\/p>\n
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\nb) T\u00ecm t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m M tr\u00ean ( C ) sao cho ti\u1ebfp tuy\u1ebfn c\u1ee7a ( C ) t\u1ea1i M song song v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng ( d ):<\/p>\n