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{"id":26679,"date":"2018-04-13T15:24:01","date_gmt":"2018-04-13T08:24:01","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=26679"},"modified":"2018-04-13T15:24:01","modified_gmt":"2018-04-13T08:24:01","slug":"de-kiem-tra-giai-tich-12-chuong-4-phan-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-chuong-4-phan-2\/","title":{"rendered":"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 ch\u01b0\u01a1ng 4 (Ph\u1ea7n 2)"},"content":{"rendered":"

C\u00e2u 7:<\/b>\u00a0Cho ca\u0301c s\u00f4\u0301 ph\u01b0\u0301c z1<\/sub>\u00a0= 1 + i, z2<\/sub>\u00a0= 1 – i, z3<\/sub>\u00a0= 2 + 3i . Gia\u0301 tri\u0323 cu\u0309a bi\u00ea\u0309u th\u01b0\u0301c T = |z1<\/sub>z2<\/sub>\u00a0+ z2<\/sub>z3<\/sub>\u00a0+ z3<\/sub>z1<\/sub>| la\u0300<\/p>\n

A. 6\u00a0\u00a0\u00a0B. 12\u00a0\u00a0\u00a0C. 6\u221a2\u00a0\u00a0\u00a0D. 10.<\/p>\n

C\u00e2u 8:<\/b>\u00a0Nghi\u0323ch \u0111a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = 4 + 3i la\u0300<\/p>\n

\"\"<\/p>\n

C\u00e2u 9:<\/b><\/p>\n

\"\"<\/p>\n

A. z = 3 – i \u00a0\u00a0\u00a0B. z = 3 + i\u00a0\u00a0\u00a0C. z = -3 + i\u00a0\u00a0\u00a0D. z = -3 – i<\/p>\n

C\u00e2u 10:<\/b>\u00a0Ca\u0301c s\u00f4\u0301 th\u01b0\u0323c x, y tho\u0309a ma\u0303n \u0111\u0103\u0309ng th\u01b0\u0301c x(3 + 5i) – y(1 + 2i) = 9 + 16i . Gia\u0301 tri\u0323 bi\u00ea\u0309u th\u01b0\u0301c T = |x – y| la\u0300<\/p>\n

A. 0 \u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0C. 3\u00a0\u00a0\u00a0D. 5.<\/p>\n

C\u00e2u 11:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n (1 + 2i)2<\/sup>.z +\u00a0z\u2212<\/i>\u00a0= 4i – 20. M\u00f4\u0111un cu\u0309a z la\u0300<\/p>\n

A. 4\u00a0\u00a0\u00a0B. 5 \u00a0\u00a0\u00a0C. 6\u00a0\u00a0\u00a0D. 10<\/p>\n

C\u00e2u 12:<\/b>\u00a0Ph\u01b0\u01a1ng tri\u0300nh z2<\/sup>\u00a0+ az + b = 0 nh\u00e2\u0323n z = 1 – 2i la\u0300m nghi\u00ea\u0323m. Khi \u0111o\u0301 a + b b\u0103\u0300ng<\/p>\n

A. 3 \u00a0\u00a0\u00a0B. 4\u00a0\u00a0\u00a0C. 5\u00a0\u00a0\u00a0D. 6.<\/p>\n

C\u00e2u 13:<\/b>\u00a0Ph\u01b0\u01a1ng tri\u0300nh z2<\/sup>\u00a0+ 1 = 2\u221a2i co\u0301 ca\u0301c nghi\u00ea\u0323m la\u0300 z1<\/sub>, z2<\/sub>\u00a0. Ti\u0301nh T = |z1<\/sub>| + |z2<\/sub>|<\/p>\n

A. 2\u00a0\u00a0\u00a0B. 2\u221a2\u00a0\u00a0\u00a0C. 2\u221a3\u00a0\u00a0\u00a0D. 12<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
7-C<\/td>\n8-D<\/td>\n9-A<\/td>\n10-D<\/td>\n11-B<\/td>\n12-A<\/td>\n13-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 7:<\/b><\/p>\n

Ta c\u00f3: z1<\/sub>z2<\/sub>\u00a0+ z2<\/sub>z3<\/sub>\u00a0+ z3<\/sub>z1<\/sub>\u00a0= z1<\/sub>z2<\/sub>\u00a0+ z3<\/sub>(z1<\/sub>\u00a0+ z2<\/sub>) = 1 – i2<\/sup>\u00a0+ 2(2 + 3i) = 6 + 6i<\/p>\n

Do \u0111\u00f3: T = | z1<\/sub>z2<\/sub>\u00a0+ z2<\/sub>z3<\/sub>\u00a0+ z3<\/sub>z1<\/sub>\u00a0| = 6\u221a2<\/p>\n

C\u00e2u 8:<\/b><\/p>\n

Ngh\u1ecbch \u0111\u1ea3o c\u1ee7a s\u1ed1 ph\u1ee9c z = 4 + 3i l\u00e0<\/p>\n

\"\"<\/p>\n

C\u00e2u 9:<\/b><\/p>\n

Ta c\u00f3<\/p>\n

\"\"<\/p>\n

C\u00e2u 10:<\/b><\/p>\n

Ta c\u00f3: x(3 + 5i) – y(1 + 2i) = 9 + 16i <=> (3x – y) + (5x – 2y) = 9 + 16i<\/p>\n

\"\"<\/p>\n

V\u1eady: T = |x – y| = 5<\/p>\n

C\u00e2u 11:<\/b><\/p>\n

\u0110\u1eb7t a + bi(a, b \u2208R). Ta c\u00f3:<\/p>\n

(1 + 2i)2<\/sup>z = (1 + 2i – 4)(a + bi) = -3a – 3bi + 4ai – 4b = -3a – 4b + (4a – 3b)i<\/p>\n

Do \u0111\u00f3: (1 + 2i)2<\/sup>.z +\u00a0a\u2212<\/i>\u00a0= 4i – 20 <=> -3a – 4b + (4a – 3b)i + a – bi = 4i – 20<\/p>\n

<=> -2a – 4b + (4a – 4b)i = 4i – 20<\/p>\n

\"\"<\/p>\n

C\u00e2u 12:<\/b><\/p>\n

Ta c\u00f3 z = 1 – 2i l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho n\u00ean:<\/p>\n

(1 – 2i)2<\/sup>\u00a0+ a(1 – 2i) + b = 0 <=> (a + b – 3) – (2a + 4)i = 0<\/p>\n

\"\"<\/p>\n

V\u1eady: a + b = -2 + 5 = 3<\/p>\n

C\u00e2u 13:<\/b><\/p>\n

Ta c\u00f3: z2<\/sup>\u00a0= -1 + 2\u221a2i = 1 + 2\u221a2i + 2i2<\/sup>\u00a0= (1 + \u221a2i)2<\/sup>\u00a0<=> z1,2<\/sub>\u00a0= \u00b1(1 + \u221a2i)<\/p>\n

\"\"<\/p>\n

Ch\u00fa \u00fd. C\u00f3 th\u1ec3 \u0111\u1eb7t z = a + bi(a,b \u2208R). Ta c\u00f3: z2<\/sup>\u00a0= a2<\/sup>\u00a0– b2<\/sup>\u00a0+ 2abi. T\u1eeb gi\u1ea3 thi\u1ebft ta c\u00f3 :<\/p>\n

\"\"<\/p>\n

T\u1eeb \u0111\u00f3 suy ra a = 1, b = \u221a2 ho\u1eb7c a = -1, b = -\u221a2<\/p>\n

Do \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m: z1,2<\/sub>\u00a0= \u00b1(1 + *=\u221a2i)<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 7:\u00a0Cho ca\u0301c s\u00f4\u0301 ph\u01b0\u0301c z1\u00a0= 1 + i, z2\u00a0= 1 – i, z3\u00a0= 2 + 3i . Gia\u0301 tri\u0323 cu\u0309a bi\u00ea\u0309u th\u01b0\u0301c T = |z1z2\u00a0+ z2z3\u00a0+ z3z1| la\u0300 A. 6\u00a0\u00a0\u00a0B. 12\u00a0\u00a0\u00a0C. 6\u221a2\u00a0\u00a0\u00a0D. 10. C\u00e2u 8:\u00a0Nghi\u0323ch \u0111a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = 4 + 3i la\u0300 C\u00e2u 9: A. z = 3 – i \u00a0\u00a0\u00a0B. […]<\/p>\n","protected":false},"author":3,"featured_media":26680,"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 Gi\u1ea3i t\u00edch 12 ch\u01b0\u01a1ng 4 (Ph\u1ea7n 2)<\/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-giai-tich-12-chuong-4-phan-2\/\" \/>\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 Gi\u1ea3i t\u00edch 12 ch\u01b0\u01a1ng 4 (Ph\u1ea7n 2)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 7:\u00a0Cho ca\u0301c s\u00f4\u0301 ph\u01b0\u0301c z1\u00a0= 1 + i, z2\u00a0= 1 – i, z3\u00a0= 2 + 3i . Gia\u0301 tri\u0323 cu\u0309a bi\u00ea\u0309u th\u01b0\u0301c T = |z1z2\u00a0+ z2z3\u00a0+ z3z1| la\u0300 A. 6\u00a0\u00a0\u00a0B. 12\u00a0\u00a0\u00a0C. 6\u221a2\u00a0\u00a0\u00a0D. 10. C\u00e2u 8:\u00a0Nghi\u0323ch \u0111a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = 4 + 3i la\u0300 C\u00e2u 9: A. z = 3 – i \u00a0\u00a0\u00a0B. […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-chuong-4-phan-2\/\" \/>\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=\"2018-04-13T08:24:01+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/1-181.png\" \/>\n\t<meta property=\"og:image:width\" content=\"255\" \/>\n\t<meta property=\"og:image:height\" content=\"135\" \/>\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-giai-tich-12-chuong-4-phan-2\/\",\"url\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-chuong-4-phan-2\/\",\"name\":\"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 ch\u01b0\u01a1ng 4 (Ph\u1ea7n 2)\",\"isPartOf\":{\"@id\":\"https:\/\/lop12.edu.vn\/#website\"},\"datePublished\":\"2018-04-13T08:24:01+00:00\",\"dateModified\":\"2018-04-13T08:24:01+00:00\",\"author\":{\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\"},\"breadcrumb\":{\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-chuong-4-phan-2\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-chuong-4-phan-2\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-chuong-4-phan-2\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/lop12.edu.vn\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 ch\u01b0\u01a1ng 4 (Ph\u1ea7n 2)\"}]},{\"@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 Gi\u1ea3i t\u00edch 12 ch\u01b0\u01a1ng 4 (Ph\u1ea7n 2)","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-giai-tich-12-chuong-4-phan-2\/","og_locale":"en_US","og_type":"article","og_title":"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 ch\u01b0\u01a1ng 4 (Ph\u1ea7n 2)","og_description":"C\u00e2u 7:\u00a0Cho ca\u0301c s\u00f4\u0301 ph\u01b0\u0301c z1\u00a0= 1 + i, z2\u00a0= 1 – i, z3\u00a0= 2 + 3i . Gia\u0301 tri\u0323 cu\u0309a bi\u00ea\u0309u th\u01b0\u0301c T = |z1z2\u00a0+ z2z3\u00a0+ z3z1| la\u0300 A. 6\u00a0\u00a0\u00a0B. 12\u00a0\u00a0\u00a0C. 6\u221a2\u00a0\u00a0\u00a0D. 10. C\u00e2u 8:\u00a0Nghi\u0323ch \u0111a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = 4 + 3i la\u0300 C\u00e2u 9: A. z = 3 – i \u00a0\u00a0\u00a0B. 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