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 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\n\n
\n 7-C<\/td>\n 8-D<\/td>\n 9-A<\/td>\n 10-D<\/td>\n 11-B<\/td>\n 12-A<\/td>\n 13-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n