D\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0110\u1ec1 thi th\u1eed THPT qu\u1ed1c gia M\u00f4n To\u00e1n n\u0103m 2016 (S\u01b0u t\u1ea7m). 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!<\/em><\/p>\n

C\u00e2u 1(2.0 \u0111i\u1ec3m)<\/strong><\/p>\n

Cho h\u00e0m s\u1ed1 y= \u2212x<\/em>^{3 <\/sup>+ 3x<\/em>2 <\/sup>+ (m <\/em>\u2212 1)x <\/em>\u2212 1.<\/p>\n}

a. Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho khi m <\/em>= 1.<\/p>\n

b. T\u00ecm t\u1ea5t c\u1ea3 c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a m <\/em>\u0111\u1ec3 h\u00e0m s\u1ed1 \u0111\u00e3 cho \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (0;3).<\/p>\n

C\u00e2u 2 (1,0 \u0111i\u1ec3m).<\/strong><\/p>\n

a. Cho cung \u03b1 <\/em>th\u1ecfa m\u00e3n tan\u03b1 <\/em>= 2.<\/p>\n

<\/a><\/p>\n

b. T\u00ecm m\u00f4\u0111un c\u1ee7a s\u1ed1 ph\u1ee9c:<\/p>\n

<\/a><\/p>\n

C\u00e2u 3 (0,5 \u0111i\u1ec3m)<\/strong>. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n

<\/a><\/p>\n

C\u00e2u 4 (1,0 \u0111i\u1ec3m)<\/strong>. Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n

<\/a><\/p>\n

C\u00e2u 5 (1,0 \u0111i\u1ec3m)<\/strong>. T\u00ednh t\u00edch ph\u00e2n:<\/p>\n

<\/a><\/p>\n

C\u00e2u 6 (1,0 \u0111i\u1ec3m)<\/strong>. Cho h\u00ecnh ch\u00f3p S.ABCD <\/em>c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh thoi, tam gi\u00e1c SAB <\/em>\u0111\u1ec1u v\u00e0 n\u1eb1m trong m\u1eb7t ph\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABCD<\/em>). Bi\u1ebft AC <\/em>= 2a,BD <\/em>= 4a<\/em>, t\u00ednh theo a <\/em>th\u1ec3 t\u00edch kh\u1ed1i ch\u00f3p S.ABCD <\/em>v\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng AD <\/em>v\u00e0 SC<\/em>.<\/p>\n

C\u00e2u 7 (1,0 \u0111i\u1ec3m).<\/strong> Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxy<\/em>, cho tam gi\u00e1c ABC <\/em>c\u00f3 \u0111\u1ec9nh A<\/em>(1;2). \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c trong v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng ch\u1ee9a trung tuy\u1ebfn k\u1ebb t\u1eeb B <\/em>l\u1ea7n l\u01b0\u1ee3t c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh 2x <\/em>\u2212 y <\/em>+ 5 = 0 v\u00e0 7x <\/em>\u2212 y <\/em>+ 15 = 0. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC<\/em>.<\/p>\n

C\u00e2u 8 (1,0 \u0111i\u1ec3m)<\/strong>. Trong kh\u00f4ng gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz<\/em>, cho m\u1eb7t ph\u1eb3ng (P<\/em>) : 3x<\/em>+5y<\/em>\u2212z<\/em>\u22122 = 0 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d)<\/p>\n

<\/a>. T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m M <\/em>c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d <\/em>v\u00e0 m\u1eb7t ph\u1eb3ng (P<\/em>). Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (Q<\/em>) ch\u1ee9a d <\/em>v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi (P<\/em>).<\/p>\n

C\u00e2u 9 (0,5 \u0111i\u1ec3m)<\/strong>. C\u00f3 hai c\u00e1i h\u1ed9p A <\/em>v\u00e0 B <\/em>\u0111\u1ef1ng c\u00e1c c\u00e2y vi\u1ebft. H\u1ed9p A <\/em>g\u1ed3m 5 c\u00e2y vi\u1ebft m\u00e0u \u0111\u1ecf v\u00e0 6 c\u00e2y vi\u1ebft m\u00e0u xanh. H\u1ed9p B <\/em>g\u1ed3m 7 c\u00e2y vi\u1ebft m\u00e0u \u0111\u1ecf v\u00e0 8 c\u00e2y vi\u1ebft m\u00e0u xanh. L\u1ea5y ng\u1eabu nhi\u00ean c\u00f9ng m\u1ed9t l\u00fac t\u1eeb m\u1ed7i h\u1ed9p ra m\u1ed9t c\u00e2y vi\u1ebft. T\u00ednh x\u00e1c su\u1ea5t sao cho hai c\u00e2y vi\u1ebft \u0111\u01b0\u1ee3c l\u1ea5y ra c\u00f3 c\u00f9ng m\u00e0u.<\/p>\n

C\u00e2u 10 (1,0 \u0111i\u1ec3m)<\/strong>. Cho c\u00e1c s\u1ed1 th\u1ef1c d\u01b0\u01a1ng a,b,c <\/em>th\u1ecfa m\u00e3n 9(a<\/em>^{4 <\/sup>+ b<\/em>4 <\/sup>+ c<\/em>4<\/sup>)\u221225(a<\/em>2 <\/sup>+ b<\/em>2 <\/sup>+ c<\/em>2<\/sup>)+ 48 = 0. T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c:<\/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

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

D\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0110\u1ec1 thi th\u1eed THPT qu\u1ed1c gia M\u00f4n To\u00e1n n\u0103m 2016 (S\u01b0u t\u1ea7m). 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! C\u00e2u 1(2.0 \u0111i\u1ec3m) Cho h\u00e0m s\u1ed1 y= \u2212x3 + 3×2 + (m \u2212 1)x \u2212 1. a. Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean […]<\/p>\n","protected":false},"author":2,"featured_media":18291,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[28],"tags":[],"yoast_head":"\n