C\u00e2u 8:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n: i.z\u2212<\/i>\u00a0+ z = 2 + 2i v\u00e0 z.z\u2212<\/i>\u00a0= 2. Khi \u0111o\u0301 z2<\/sup>\u00a0b\u0103\u0300ng:<\/p>\n A. 2 \u00a0\u00a0\u00a0B. 4\u00a0\u00a0\u00a0C. \u2013 2i \u00a0\u00a0\u00a0D. 2i.<\/p>\n C\u00e2u 9:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n (1 + i)(z – i) + 2z = 2i. M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c:<\/p>\n <\/p>\n A. 2 \u00a0\u00a0\u00a0B. 4\u00a0\u00a0\u00a0C. \u221a10\u00a0\u00a0\u00a0D. 10<\/p>\n C\u00e2u 10:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n<\/p>\n <\/p>\n Khi \u0111o\u0301 m\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c w = 1 + z + z2<\/sup>\u00a0la\u0300<\/p>\n A. 5\u00a0\u00a0\u00a0B. \u221a13 \u00a0\u00a0\u00a0C. 13 \u00a0\u00a0\u00a0D. \u221a5<\/p>\n C\u00e2u 11:<\/b>\u00a0Ph\u01b0\u01a1ng tri\u0300nh z2<\/sup>\u00a0– 2z + 3 = 0 co\u0301 ca\u0301c nghi\u00ea\u0323m la\u0300<\/p>\n A. 2\u00b12\u221a2i\u00a0\u00a0\u00a0B. -2\u00b12\u221a2i \u00a0\u00a0\u00a0 C. -1\u00b12\u221a2i\u00a0\u00a0\u00a0D. 1\u00b12\u221a2i<\/p>\n C\u00e2u 12:<\/b>\u00a0Ph\u01b0\u01a1ng tri\u0300nh z4<\/sup>\u00a0– 2z2<\/sup>\u00a0– 3 = 0 co\u0301 4 nghi\u00ea\u0323m ph\u01b0\u0301c z1<\/sub>, z2<\/sub>, z3<\/sub>, z4<\/sub>. Gia\u0301 tri\u0323 bi\u00ea\u0309u th\u01b0\u0301c T = |z1<\/sub>|2<\/sup>\u00a0+ |z2<\/sub>|2<\/sup>\u00a0+ |z3<\/sub>|2<\/sup>\u00a0+ |z4<\/sub>|2<\/sup>b\u0103\u0300ng<\/p>\n A. 4 \u00a0\u00a0\u00a0B. 8\u00a0\u00a0\u00a0C. 2\u221a3\u00a0\u00a0\u00a0D. 2 + 2\u221a3<\/p>\n C\u00e2u 13:<\/b>\u00a0T\u00e2\u0323p h\u01a1\u0323p ca\u0301c \u0111i\u00ea\u0309m bi\u00ea\u0309u di\u00ea\u0303n s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n |z – 2i| = 4 la\u0300<\/p>\n A. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(1; -2) ba\u0301n ki\u0301nh R = 4<\/p>\n B. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(1; 2) ba\u0301n ki\u0301nh R = 4<\/p>\n C. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(0; 2) ba\u0301n ki\u0301nh R = 4<\/p>\n D. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(0; -2) ba\u0301n ki\u0301nh R = 4<\/p>\n C\u00e2u 14:<\/b>\u00a0T\u00e2\u0323p h\u01a1\u0323p ca\u0301c \u0111i\u00ea\u0309m bi\u00ea\u0309u di\u00ea\u0303n s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n |z\u2212<\/i>\u00a0+ 3 – 2i| = 4 la\u0300<\/p>\n A. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(3; 2) ba\u0301n ki\u0301nh R = 4<\/p>\n B. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(3; -2) ba\u0301n ki\u0301nh R = 4<\/p>\n C. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(-3; 2) ba\u0301n ki\u0301nh R = 4<\/p>\n D. \u0110\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m I(-3; -2) ba\u0301n ki\u0301nh R = 4<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 8:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3:\u00a0z\u2212<\/i>\u00a0= a – bi v\u00e0 z.z\u2212<\/i>\u00a0= a2<\/sup>\u00a0+ b2<\/sup>\u00a0= 2(1)<\/p>\n Ta c\u00f3: i.z\u2212<\/i>\u00a0+ z = 2 + 2i <=> i(a – bi) + a + bi = 2 + 2i<\/p>\n <=> a + b + (a + b)i = 2 + 2i <=> a + b = 2 (2)<\/p>\n T\u1eeb (1) v\u00e0 (2) suy ra a = b = 1. Suy ra z=1+i<\/p>\n V\u1eady z2<\/sup>\u00a0= (1 + i)2<\/sup>\u00a0= 1 + 2i – 1 = 2i<\/p>\n C\u00e2u 9:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3 :<\/p>\n (1 + i)(z – i) = (1 + i)[a + (b – 1)i] = a – b + 1 + (a + b – 1)i<\/p>\n T\u1eeb gi\u1ea3 thi\u1ebft ta c\u00f3: (1 + i)(z – 1) + 2z = 2i<\/p>\n <=> a – b + 1 + (a + b – 1)i + 2(a + bi) = 2i <=> (3a – b + 1) + (a + 3b – 1)i = 2i<\/p>\n <\/p>\n Suy ra z = 1 v\u00e0<\/p>\n <\/p>\n C\u00e2u 10:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3<\/p>\n <\/p>\n <=> 5a – 5(b – 1)i = (2 – i)(a + 1 + bi)<\/p>\n <=> 3a – b – 2 + (a – 7b + 6)i = 0<\/p>\n <\/p>\n Suy ra z = 1 + i v\u00e0 w = 1 + (1 + i) + (1 + i)2<\/sup>\u00a0= 2 + 3i.<\/p>\n V\u1eady: |w| = \u221a(4 + 9) = \u221a13<\/p>\n C\u00e2u 11:<\/b><\/p>\n Ta c\u00f3: \u0394’ = 12<\/sup>\u00a0– 3 = -2 = 2i2<\/sup>. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m: z1,2<\/sub>\u00a0= 1 \u00b1 2i<\/p>\n C\u00e2u 12:<\/b><\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi: z2<\/sup>\u00a0= -1 = i2<\/sup>\u00a0ho\u1eb7c z2<\/sup>\u00a0= 3. C\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: z1<\/sub>\u00a0= i, z2<\/sub>\u00a0= -i, z3<\/sub>\u00a0= \u221a3, z4<\/sub>\u00a0= -\u221a-3.<\/p>\n V\u1eady T = 1 + 1 + 3 + 3 = 8<\/p>\n C\u00e2u 13:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3:<\/p>\n |z – 2i| = 4 <=> |a + (b – 2)i| = 4<\/p>\n <\/p>\n V\u1eady t\u1eadp c\u00e1c \u0111i\u1ec3m bi\u1ec3u di\u1ec5n s\u1ed1 ph\u1ee9c z l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m I(0 ;2), b\u00e1n k\u00ednh R = 4<\/p>\n C\u00e2u 14:<\/b><\/p>\n \u0110\u1eb7t z = a + bi(a, b \u2208 R). Ta c\u00f3: |z\u2212<\/i>\u00a0+ 3 – 2i| = 4 <=> |a – bi + 3 – 2i| = 4<\/p>\n <=> |(a + 3) – (b + 2)i| = 4<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 8:\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n: i.z\u2212\u00a0+ z = 2 + 2i v\u00e0 z.z\u2212\u00a0= 2. Khi \u0111o\u0301 z2\u00a0b\u0103\u0300ng: A. 2 \u00a0\u00a0\u00a0B. 4\u00a0\u00a0\u00a0C. \u2013 2i \u00a0\u00a0\u00a0D. 2i. C\u00e2u 9:\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n (1 + i)(z – i) + 2z = 2i. M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c: A. 2 \u00a0\u00a0\u00a0B. 4\u00a0\u00a0\u00a0C. \u221a10\u00a0\u00a0\u00a0D. 10 C\u00e2u 10:\u00a0Cho […]<\/p>\n","protected":false},"author":3,"featured_media":26694,"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 8-D<\/td>\n 9-C<\/td>\n 10-B<\/td>\n 11-D<\/td>\n 12-B<\/td>\n 13-C<\/td>\n 14-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n