\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT chuy\u00ean V\u0129nh ph\u00fac t\u1ed5 ch\u1ee9c thi th\u1eed l\u1ea7n 1. \u0110\u1ec1 thi \u0111\u01b0\u1ee3c ra theo c\u1ea5u tr\u00fac m\u1edbi nh\u1ea5t c\u1ee7a B\u1ed9 GD&\u0110T.<\/strong><\/h2>\n\nXem th\u00eam: \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n To\u00e1n<\/p>\n<\/div>\n
C\u00e2u 1 (1,0 \u0111i\u1ec3m).<\/strong> Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = x<\/p>\n3<\/dd>\n\u2013 3x<\/p>\n
2<\/dd>\n+ 2<\/p>\n
C\u00e2u 2 (1,0 \u0111i\u1ec3m).<\/strong>T\u00ecm c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1: y = x \u2013 sin2x + 2<\/p>\nC\u00e2u 3 (1,0 \u0111i\u1ec3m)<\/strong><\/p>\na) Cho tan \u03b1 = 3. T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c M = (3sin\u03b1 \u2013 2 cos\u03b1)\/(5sin<\/p>\n
3<\/dd>\n\u03b1 + 4cos<\/p>\n
3<\/dd>\n\u03b1)<\/p>\n
b) T\u00ednh gi\u1edbi h\u1ea1n: L = lim<\/p>\nx<\/dd>\n<\/em><\/p>\n
\u2192<\/dd>\n3<\/dd>\n(x-\u221a4x-3)\/(x<\/p>\n
2<\/dd>\n\u2013 9)<\/p>\n
C\u00e2u 4 (1,0 \u0111i\u1ec3m).<\/strong>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: 3sin<\/p>\n2<\/dd>\nx \u2013 4sinxcosx + 5 cos<\/p>\n
2<\/dd>\nx = 2<\/p>\n
C\u00e2u 5 (1,0 \u0111i\u1ec3m)<\/strong><\/p>\na) T\u00ecm h\u1ec7 s\u1ed1 c\u1ee7a x10 trong khai tri\u1ec3n c\u1ee7a bi\u1ec3u th\u1ee9c: (3x<\/p>\n
3<\/dd>\n\u2013 2\/x<\/p>\n
2<\/dd>\n)<\/p>\n
5<\/dd>\n<\/p>\nb) M\u1ed9t h\u1ed9p ch\u1ee9a 20 qu\u1ea3 c\u1ea7u gi\u1ed1ng nhau g\u1ed3m 12 qu\u1ea3 \u0111\u1ecf v\u00e0 8 qu\u1ea3 xanh. L\u1ea5y ng\u1eabu nhi\u00ean (\u0111\u1ed3ng th\u1eddi) 3 qu\u1ea3. T\u00ednh x\u00e1c su\u1ea5t \u0111\u1ec3 c\u00f3 \u00edt nh\u1ea5t m\u1ed9t qu\u1ea3 c\u1ea7u m\u00e0u xanh.<\/p>\n
C\u00e2u 6 (1,0 \u0111i\u1ec3m).<\/strong>Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 (Oxy), cho h\u00ecnh b\u00ecnh h\u00e0nh ABCD c\u00f3 hai \u0111\u1ec9nh A(-2;-1), D(5;0) v\u00e0 c\u00f3 t\u00e2m I(2;1). H\u00e3y x\u00e1c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 hai \u0111\u1ec9nh B, C v\u00e0 g\u00f3c nh\u1ecdn h\u1ee3p b\u1edfi hai \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a h\u00ecnh b\u00ecnh h\u00e0nh \u0111\u00e3 cho.<\/p>\nC\u00e2u 7 (1,0 \u0111i\u1ec3m)<\/strong><\/p>\nCho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u1ea5y ABC l\u00e0 tam gi\u00e1c vu\u00f4ng t\u1ea1i A, m\u1eb7t b\u00ean SAB l\u00e0 tam gi\u00e1c \u0111\u1ec1u v\u00e0 n\u1eb1m trong m\u1eb7t ph\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABC), g\u1ecdi M l\u00e0 \u0111i\u1ec3m thu\u1ed9c c\u1ea1nh SC sao cho MC = 2MS. Bi\u1ebft AB = 3, BC = 3\u221a3, t\u00ednh th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i ch\u00f3p S.ABC v\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng AC v\u00e0 BM.<\/p>\n
C\u00e2u 8. (1,0 \u0111i\u1ec3m).<\/strong>Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 (Oxy), cho tam gi\u00e1c ABC ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m J(2;1). Bi\u1ebft \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh A c\u1ee7a tam gi\u00e1c ABC c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 2x + y \u2013 10 = 0 v\u00e0 D(2;-4) l\u00e0 giao \u0111i\u1ec3m th\u1ee9 hai c\u1ee7a AJ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c ABC. T\u00ecm t\u1ecda \u0111\u1ed9 c\u00e1c \u0111\u1ec9nh tam gi\u00e1c ABC bi\u1ebft B c\u00f3 ho\u00e0nh \u0111\u1ed9 \u00e2m v\u00e0 B thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x + y + 7 = 0.<\/p>\n![](\"http:\/\/img.toanhoc247.com\/picture\/2015\/1012\/de-thi-thu-thppt-2016-toan-chuyenvp-2.png\")
<\/p>\n<\/p>\n
T\u1ea5t c\u1ea3 n\u1ed9i dung b\u00e0i vi\u1ebft. C\u00e1c em h\u00e3y xem th\u00eam v\u00e0 t\u1ea3i file chi ti\u1ebft t\u1ea1i \u0111\u00e2y:<\/strong>Download<\/p>\n","protected":false},"excerpt":{"rendered":"\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT chuy\u00ean V\u0129nh ph\u00fac t\u1ed5 ch\u1ee9c thi th\u1eed l\u1ea7n 1. \u0110\u1ec1 thi \u0111\u01b0\u1ee3c ra theo c\u1ea5u tr\u00fac m\u1edbi nh\u1ea5t c\u1ee7a B\u1ed9 GD&\u0110T. Xem th\u00eam: \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n To\u00e1n C\u00e2u 1 (1,0 \u0111i\u1ec3m). Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[28],"tags":[],"yoast_head":"\n
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Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 […]","og_url":"https:\/\/lop12.edu.vn\/de-thi-thu-thpt-quoc-gia-mon-toan-2016-lan-1-chuyen-vinh-phuc\/","og_site_name":"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam","article_author":"www.facebook.com\/lop12.edu.vn\/","article_published_time":"2016-03-19T22:50:45+00:00","article_modified_time":"2016-03-20T15:52:28+00:00","og_image":[{"url":"http:\/\/img.toanhoc247.com\/picture\/2015\/1012\/de-thi-thu-thppt-2016-toan-chuyenvp-2.png"}],"author":"Lop12","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Lop12","Est. reading time":"3 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/lop12.edu.vn\/de-thi-thu-thpt-quoc-gia-mon-toan-2016-lan-1-chuyen-vinh-phuc\/","url":"https:\/\/lop12.edu.vn\/de-thi-thu-thpt-quoc-gia-mon-toan-2016-lan-1-chuyen-vinh-phuc\/","name":"\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n To\u00e1n 2016 L\u1ea7n 1 chuy\u00ean V\u0129nh Ph\u00fac","isPartOf":{"@id":"https:\/\/lop12.edu.vn\/#website"},"datePublished":"2016-03-19T22:50:45+00:00","dateModified":"2016-03-20T15:52:28+00:00","author":{"@id":"https:\/\/lop12.edu.vn\/#\/schema\/person\/8219caa3d91a087a4ce81b8b5ac734b5"},"breadcrumb":{"@id":"https:\/\/lop12.edu.vn\/de-thi-thu-thpt-quoc-gia-mon-toan-2016-lan-1-chuyen-vinh-phuc\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/lop12.edu.vn\/de-thi-thu-thpt-quoc-gia-mon-toan-2016-lan-1-chuyen-vinh-phuc\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/lop12.edu.vn\/de-thi-thu-thpt-quoc-gia-mon-toan-2016-lan-1-chuyen-vinh-phuc\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/lop12.edu.vn\/"},{"@type":"ListItem","position":2,"name":"\u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n To\u00e1n 2016 L\u1ea7n 1 chuy\u00ean V\u0129nh Ph\u00fac"}]},{"@type":"WebSite","@id":"https:\/\/lop12.edu.vn\/#website","url":"https:\/\/lop12.edu.vn\/","name":"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/lop12.edu.vn\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/lop12.edu.vn\/#\/schema\/person\/8219caa3d91a087a4ce81b8b5ac734b5","name":"Lop12","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/lop12.edu.vn\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/65b2897b6a31b932be1276ba7fc18d41?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/65b2897b6a31b932be1276ba7fc18d41?s=96&d=mm&r=g","caption":"Lop12"},"sameAs":["http:\/\/www.lop12.edu.vn","www.facebook.com\/lop12.edu.vn\/"],"url":"https:\/\/lop12.edu.vn\/author\/hungld\/"}]}},"_links":{"self":[{"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/posts\/1481"}],"collection":[{"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/comments?post=1481"}],"version-history":[{"count":0,"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/posts\/1481\/revisions"}],"wp:attachment":[{"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/media?parent=1481"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/categories?post=1481"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/lop12.edu.vn\/wp-json\/wp\/v2\/tags?post=1481"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Xem th\u00eam: \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n To\u00e1n<\/p>\n<\/div>\n
C\u00e2u 1 (1,0 \u0111i\u1ec3m).<\/strong> Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = x<\/p>\n \u2013 3x<\/p>\n + 2<\/p>\n C\u00e2u 2 (1,0 \u0111i\u1ec3m).<\/strong>T\u00ecm c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1: y = x \u2013 sin2x + 2<\/p>\n C\u00e2u 3 (1,0 \u0111i\u1ec3m)<\/strong><\/p>\n a) Cho tan \u03b1 = 3. T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c M = (3sin\u03b1 \u2013 2 cos\u03b1)\/(5sin<\/p>\n \u03b1 + 4cos<\/p>\n \u03b1)<\/p>\n b) T\u00ednh gi\u1edbi h\u1ea1n: L = lim<\/p>\n <\/em><\/p>\n (x-\u221a4x-3)\/(x<\/p>\n \u2013 9)<\/p>\n C\u00e2u 4 (1,0 \u0111i\u1ec3m).<\/strong>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: 3sin<\/p>\n x \u2013 4sinxcosx + 5 cos<\/p>\n x = 2<\/p>\n C\u00e2u 5 (1,0 \u0111i\u1ec3m)<\/strong><\/p>\n a) T\u00ecm h\u1ec7 s\u1ed1 c\u1ee7a x10 trong khai tri\u1ec3n c\u1ee7a bi\u1ec3u th\u1ee9c: (3x<\/p>\n \u2013 2\/x<\/p>\n )<\/p>\n b) M\u1ed9t h\u1ed9p ch\u1ee9a 20 qu\u1ea3 c\u1ea7u gi\u1ed1ng nhau g\u1ed3m 12 qu\u1ea3 \u0111\u1ecf v\u00e0 8 qu\u1ea3 xanh. L\u1ea5y ng\u1eabu nhi\u00ean (\u0111\u1ed3ng th\u1eddi) 3 qu\u1ea3. T\u00ednh x\u00e1c su\u1ea5t \u0111\u1ec3 c\u00f3 \u00edt nh\u1ea5t m\u1ed9t qu\u1ea3 c\u1ea7u m\u00e0u xanh.<\/p>\n C\u00e2u 6 (1,0 \u0111i\u1ec3m).<\/strong>Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 (Oxy), cho h\u00ecnh b\u00ecnh h\u00e0nh ABCD c\u00f3 hai \u0111\u1ec9nh A(-2;-1), D(5;0) v\u00e0 c\u00f3 t\u00e2m I(2;1). H\u00e3y x\u00e1c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 hai \u0111\u1ec9nh B, C v\u00e0 g\u00f3c nh\u1ecdn h\u1ee3p b\u1edfi hai \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a h\u00ecnh b\u00ecnh h\u00e0nh \u0111\u00e3 cho.<\/p>\n C\u00e2u 7 (1,0 \u0111i\u1ec3m)<\/strong><\/p>\n Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u1ea5y ABC l\u00e0 tam gi\u00e1c vu\u00f4ng t\u1ea1i A, m\u1eb7t b\u00ean SAB l\u00e0 tam gi\u00e1c \u0111\u1ec1u v\u00e0 n\u1eb1m trong m\u1eb7t ph\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABC), g\u1ecdi M l\u00e0 \u0111i\u1ec3m thu\u1ed9c c\u1ea1nh SC sao cho MC = 2MS. Bi\u1ebft AB = 3, BC = 3\u221a3, t\u00ednh th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i ch\u00f3p S.ABC v\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng AC v\u00e0 BM.<\/p>\n C\u00e2u 8. (1,0 \u0111i\u1ec3m).<\/strong>Trong m\u1eb7t ph\u1eb3ng v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 (Oxy), cho tam gi\u00e1c ABC ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m J(2;1). Bi\u1ebft \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh A c\u1ee7a tam gi\u00e1c ABC c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 2x + y \u2013 10 = 0 v\u00e0 D(2;-4) l\u00e0 giao \u0111i\u1ec3m th\u1ee9 hai c\u1ee7a AJ v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c ABC. T\u00ecm t\u1ecda \u0111\u1ed9 c\u00e1c \u0111\u1ec9nh tam gi\u00e1c ABC bi\u1ebft B c\u00f3 ho\u00e0nh \u0111\u1ed9 \u00e2m v\u00e0 B thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x + y + 7 = 0.<\/p>\n T\u1ea5t c\u1ea3 n\u1ed9i dung b\u00e0i vi\u1ebft. C\u00e1c em h\u00e3y xem th\u00eam v\u00e0 t\u1ea3i file chi ti\u1ebft t\u1ea1i \u0111\u00e2y:<\/strong>Download<\/p>\n","protected":false},"excerpt":{"rendered":" \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia n\u0103m 2016 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT chuy\u00ean V\u0129nh ph\u00fac t\u1ed5 ch\u1ee9c thi th\u1eed l\u1ea7n 1. \u0110\u1ec1 thi \u0111\u01b0\u1ee3c ra theo c\u1ea5u tr\u00fac m\u1edbi nh\u1ea5t c\u1ee7a B\u1ed9 GD&\u0110T. Xem th\u00eam: \u0110\u1ec1 thi th\u1eed THPT Qu\u1ed1c gia m\u00f4n To\u00e1n C\u00e2u 1 (1,0 \u0111i\u1ec3m). Kh\u1ea3o s\u00e1t s\u1ef1 bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[28],"tags":[],"yoast_head":"\n<\/p>\n<\/p>\n