C\u00e2u 1:<\/b>\u00a0Cho hai s\u00f4\u0301 ph\u01b0\u0301c z1<\/sub>\u00a0= 1 + 2i, z2<\/sub>\u00a0= 2 – 3i . Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c w = 3z1<\/sub>\u00a0– 2z2<\/sub>\u00a0la\u0300<\/p>\n A. 1 va\u0300 12\u00a0\u00a0\u00a0B. -1 va\u0300 12\u00a0\u00a0\u00a0C. \u20131 va\u0300 12i\u00a0\u00a0\u00a0D. 1 va\u0300 12i.<\/p>\n C\u00e2u 2:<\/b>\u00a0Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = (1 + \u221a3i)2<\/sup>\u00a0la\u0300<\/p>\n A. 1 va\u0300 3\u00a0\u00a0\u00a0B. 1 va\u0300 -3\u00a0\u00a0\u00a0C. -2 va\u0300 2\u221a3 \u00a0\u00a0\u00a0D. 2 va\u0300 -2\u221a3 .<\/p>\n C\u00e2u 3:<\/b>\u00a0Ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = (1 + \u221ai)3<\/sup>\u00a0la\u0300<\/p>\n A. 3\u221a3\u00a0\u00a0\u00a0B. -3\u221a3\u00a0\u00a0\u00a0C. \u2013 8i \u00a0\u00a0\u00a0D. \u20138.<\/p>\n C\u00e2u 4:<\/b>\u00a0Th\u01b0\u0323c hi\u00ea\u0323n phe\u0301p ti\u0301nh:<\/p>\n <\/p>\n ta co\u0301:<\/p>\n A. T = 3 + 4i\u00a0\u00a0\u00a0B. T = -3 + 4i\u00a0\u00a0\u00a0C. T = 3 \u2013 4i\u00a0\u00a0\u00a0D. T = -3 \u2013 4i.<\/p>\n C\u00e2u 5:<\/b>\u00a0M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n \u0111i\u00ea\u0300u ki\u00ea\u0323n z + (2 – i)z\u2212<\/i>\u00a0= 13 – 3i la\u0300<\/p>\n A. 3\u00a0\u00a0\u00a0B. 5\u00a0\u00a0\u00a0C. 17\u00a0\u00a0\u00a0D. \u221a17<\/p>\n C\u00e2u 6:<\/b>\u00a0Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n (1 – i)z – 1 + 5i = 0 la\u0300<\/p>\n A. 3 va\u0300 \u20132\u00a0\u00a0\u00a0B. 3 va\u0300 2\u00a0\u00a0\u00a0C. 3 va\u0300 \u2013 2i\u00a0\u00a0\u00a0D. 3 va\u0300 2i.<\/p>\n C\u00e2u 7:<\/b>\u00a0M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n \u0111i\u00ea\u0300u ki\u00ea\u0323n (3z –\u00a0z\u2212<\/i>)(1 + i) – 5z = 8i – 1 la\u0300<\/p>\n B. 1\u00a0\u00a0\u00a0B. 5\u00a0\u00a0\u00a0C. \u221a13 \u00a0\u00a0\u00a0 D. 13.<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n Ta c\u00f3: w = 3z1<\/sub>\u00a0– 2z2<\/sub>\u00a0= 3(1 + 2i) – 2(2 – 3i) = -1 + 2i.<\/p>\n V\u1eady ph\u1ea7n th\u1ef1c v\u00e0 ph\u1ea7n \u1ea3o c\u1ee7a w l\u00e0 -1 v\u00e0 12<\/p>\n C\u00e2u 2:<\/b><\/p>\n Ta c\u00f3: z = 1 + 2\u221a3 + 3i2<\/sup>\u00a0= -2 + 2\u221a3i<\/p>\n V\u1eady ph\u1ea7n th\u1ef1c v\u00e0 ph\u1ea7n \u1ea3o c\u1ee7a z l\u00e0 -2 v\u00e0 2\u221a3<\/p>\n C\u00e2u 3:<\/b><\/p>\n Ta c\u00f3: z = i(1 + \u221a3i)3<\/sup>\u00a0= i(1 + 3\u221a3i – 9 – 3\u221a3i) = -8i .<\/p>\n V\u1eady ph\u1ea7n \u1ea3o c\u1ee7a z l\u00e0 -8<\/p>\n C\u00e2u 4:<\/b><\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n => T = -3 + 4i<\/p>\n C\u00e2u 5:<\/b><\/p>\n M\u00f4\u0111un c\u1ee7a s\u1ed1 ph\u1ee9c z th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n z + (2 – i)z\u2212<\/i>\u00a0= 13 – 3i l\u00e0:<\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3:\u00a0z\u2212<\/i>\u00a0= a – bi v\u00e0 (2 – i)z\u2212<\/i>\u00a0= (2 – i)(a – bi) = 2a – 2bi – ai – b = 2a – b – (2b + a)i<\/p>\n Do \u0111\u00f3 : z = (2 – i)z\u2212<\/i>\u00a0= 13 – 3i <=> a + bi + 2a – b – (2b + a)i = 13 – 3i<\/p>\n <\/p>\n C\u00e2u 6:<\/b><\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n V\u1eady ph\u1ea7n th\u1ef1c v\u00e0 ph\u1ea7n \u1ea3o c\u1ee7a z l\u00e0 3 v\u00e0 -2<\/p>\n C\u00e2u 7:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3:\u00a0z\u2212<\/i>\u00a0= a – bi v\u00e0 3z –\u00a0z\u2212<\/i>\u00a0= 3(a + bi) – (a – bi) = 2a + 4bi, (3z –\u00a0z\u2212<\/i>)(1 + i) = 2a – 4b + (2a + 4b)i – 5(a + bi) = 8i – 1<\/p>\n <=> -3a – 4b + (2a – b)i = -1 + 8i<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Cho hai s\u00f4\u0301 ph\u01b0\u0301c z1\u00a0= 1 + 2i, z2\u00a0= 2 – 3i . Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c w = 3z1\u00a0– 2z2\u00a0la\u0300 A. 1 va\u0300 12\u00a0\u00a0\u00a0B. -1 va\u0300 12\u00a0\u00a0\u00a0C. \u20131 va\u0300 12i\u00a0\u00a0\u00a0D. 1 va\u0300 12i. C\u00e2u 2:\u00a0Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = (1 + \u221a3i)2\u00a0la\u0300 A. 1 […]<\/p>\n","protected":false},"author":3,"featured_media":26705,"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\n\n
\n 1-B<\/td>\n 2-C<\/td>\n 3-D<\/td>\n 4-B<\/td>\n 5-D<\/td>\n 6-A<\/td>\n 7-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n