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{"id":26583,"date":"2018-04-13T12:55:48","date_gmt":"2018-04-13T05:55:48","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=26583"},"modified":"2018-04-13T12:55:48","modified_gmt":"2018-04-13T05:55:48","slug":"de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/","title":{"rendered":"\u0110\u1ec1 ki\u1ec3m tra cu\u1ed1i n\u0103m: \u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 3)"},"content":{"rendered":"

C\u00e2u 15:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301<\/p>\n

\"\"<\/p>\n

\u0110\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 co\u0301 ti\u00ea\u0323m c\u00e2\u0323n ngang la\u0300<\/p>\n

A. y = 1\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. y = 1\/2\u00a0\u00a0\u00a0D. y = -5<\/p>\n

C\u00e2u 16:<\/b>\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh |x3<\/sup>| – 12|x| = m (v\u01a1\u0301i -1 < m < 0 ) la\u0300<\/p>\n

A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4.<\/p>\n

C\u00e2u 17:<\/b>\u00a0Cho hai s\u00f4\u0301 d\u01b0\u01a1ng a, b tho\u0309a ma\u0303n a2<\/sup>\u00a0+ b2<\/sup>\u00a0= 7ab. \u0110\u0103\u0309ng th\u01b0\u0301c na\u0300o sau \u0111\u00e2y \u0111u\u0301ng?<\/p>\n

\"\"<\/p>\n

A. 1 \u00a0\u00a0\u00a0B. 2 \u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4.<\/p>\n

C\u00e2u 18:<\/b>\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh log1\/4<\/sub>(x2<\/sup>\u00a0– x4<\/sup>) = 1 la\u0300<\/p>\n

A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4.<\/p>\n

C\u00e2u 19:<\/b>\u00a0Gia\u0309 s\u01b0\u0309 x la\u0300 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh:<\/p>\n

\"\"<\/p>\n

Khi \u0111o\u0301 ta co\u0301:<\/p>\n

A. lg(1 – x) = 1\u00a0\u00a0\u00a0B. lg(1 – x) = \u221a3<\/p>\n

C. lg(1 – x) < 1 \u00a0\u00a0\u00a0D. lg(1 – x) > \u221a3<\/p>\n

C\u00e2u 20:<\/b>\u00a0T\u00e2\u0323p nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh 32x + 1<\/sup>\u00a0– 4.3x + 1<\/sup>\u00a0+ 9 \u2264 0 la\u0300<\/p>\n

A. x \u2265 0\u00a0\u00a0\u00a0B. x \u2264 1 \u00a0\u00a0\u00a0 C. 0 \u2264 x \u2264 1 \u00a0\u00a0\u00a0D. 0 \u2264 x \u2264 2<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
15-C<\/td>\n16-D<\/td>\n17-B<\/td>\n18-B<\/td>\n19-D<\/td>\n20-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 15:<\/b><\/p>\n

Ta c\u00f3<\/p>\n

\"\"<\/p>\n

Suy ra:<\/p>\n

\"\"<\/p>\n

V\u1eady \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 ti\u1ec7m c\u1eadn ngang l\u00e0 y = 1\/2<\/p>\n

C\u00e2u 16:<\/b><\/p>\n

S\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh |x3<\/sup>| – 12|x| = m l\u00e0 s\u1ed1 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng d: y = m v\u1edbi \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = |x3<\/sup>| – 12|x|. D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb suy ra v\u1edbi -1 < m < 0 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 4 nghi\u1ec7m.<\/p>\n

C\u00e2u 17:<\/b><\/p>\n

Ta c\u00f3: a2<\/sup>\u00a0+ b2<\/sup>\u00a0= 7ab<\/p>\n

\"\"<\/p>\n

L\u1ea5y logarit c\u01a1 s\u1ed1 7 hai v\u1ebf ta c\u00f3<\/p>\n

\"\"<\/p>\n

C\u00e2u 18:<\/b><\/p>\n

Ta c\u00f3: log1\/4<\/sub>(x2<\/sup>\u00a0– x4<\/sup>) = 1<\/p>\n

\"\"<\/p>\n

V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m.<\/p>\n

C\u00e2u 19:<\/b><\/p>\n

\u0110i\u1ec1u ki\u1ec7n: -1 < x < 1. Ta c\u00f3:<\/p>\n

\"\"<\/p>\n

Do \u0111\u00f3 lg(1 – x) > \u221a3<\/p>\n

C\u00e2u 20:<\/b><\/p>\n

\u0110\u1eb7t t = 3x<\/sup>\u00a0(t > 0). B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho tr\u1edf th\u00e0nh<\/p>\n

3t2<\/sup>\u00a0– 12t + 9 \u2264 0 <=> 1 \u2264 t \u2264 3<\/p>\n

Suy ra 1 \u2264 3x<\/sup>\u00a0\u2264 3<\/p>\n

V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 0 \u2264 x \u2264 1<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 15:\u00a0Cho ha\u0300m s\u00f4\u0301 \u0110\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 co\u0301 ti\u00ea\u0323m c\u00e2\u0323n ngang la\u0300 A. y = 1\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. y = 1\/2\u00a0\u00a0\u00a0D. y = -5 C\u00e2u 16:\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh |x3| – 12|x| = m (v\u01a1\u0301i -1 < m < 0 ) la\u0300 A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4. C\u00e2u 17:\u00a0Cho hai […]<\/p>\n","protected":false},"author":3,"featured_media":26584,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1447,1446],"yoast_head":"\n\u0110\u1ec1 ki\u1ec3m tra cu\u1ed1i n\u0103m: \u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 3)<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u0110\u1ec1 ki\u1ec3m tra cu\u1ed1i n\u0103m: \u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 3)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 15:\u00a0Cho ha\u0300m s\u00f4\u0301 \u0110\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 co\u0301 ti\u00ea\u0323m c\u00e2\u0323n ngang la\u0300 A. y = 1\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. y = 1\/2\u00a0\u00a0\u00a0D. y = -5 C\u00e2u 16:\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh |x3| – 12|x| = m (v\u01a1\u0301i -1 < m < 0 ) la\u0300 A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4. C\u00e2u 17:\u00a0Cho hai […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/\" \/>\n<meta property=\"og:site_name\" content=\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam\" \/>\n<meta property=\"article:published_time\" content=\"2018-04-13T05:55:48+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/1-171.png\" \/>\n\t<meta property=\"og:image:width\" content=\"80\" \/>\n\t<meta property=\"og:image:height\" content=\"58\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"H\u00e0 Trang\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"H\u00e0 Trang\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/\",\"url\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/\",\"name\":\"\u0110\u1ec1 ki\u1ec3m tra cu\u1ed1i n\u0103m: \u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 3)\",\"isPartOf\":{\"@id\":\"https:\/\/lop12.edu.vn\/#website\"},\"datePublished\":\"2018-04-13T05:55:48+00:00\",\"dateModified\":\"2018-04-13T05:55:48+00:00\",\"author\":{\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\"},\"breadcrumb\":{\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/lop12.edu.vn\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"\u0110\u1ec1 ki\u1ec3m tra cu\u1ed1i n\u0103m: \u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 3)\"}]},{\"@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\/2da855e5d06b553ac0065157cc261a45\",\"name\":\"H\u00e0 Trang\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"caption\":\"H\u00e0 Trang\"},\"url\":\"https:\/\/lop12.edu.vn\/author\/trangvth2\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"\u0110\u1ec1 ki\u1ec3m tra cu\u1ed1i n\u0103m: \u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 3)","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/lop12.edu.vn\/de-kiem-tra-cuoi-nam-de-kiem-tra-giai-tich-12-cuoi-nam-phan-3\/","og_locale":"en_US","og_type":"article","og_title":"\u0110\u1ec1 ki\u1ec3m tra cu\u1ed1i n\u0103m: \u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 3)","og_description":"C\u00e2u 15:\u00a0Cho ha\u0300m s\u00f4\u0301 \u0110\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 co\u0301 ti\u00ea\u0323m c\u00e2\u0323n ngang la\u0300 A. y = 1\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. y = 1\/2\u00a0\u00a0\u00a0D. y = -5 C\u00e2u 16:\u00a0S\u00f4\u0301 nghi\u00ea\u0323m cu\u0309a ph\u01b0\u01a1ng tri\u0300nh |x3| – 12|x| = m (v\u01a1\u0301i -1 < m < 0 ) la\u0300 A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0C. 3 \u00a0\u00a0\u00a0D. 4. 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