C\u00e2u 1:<\/b>\u00a0Cho hai s\u00f4\u0301 ph\u01b0\u0301c z_{1<\/sub>\u00a0= 2 + 3i, z2<\/sub>\u00a0= 1 – 2i . Ti\u0300m kh\u0103\u0309ng \u0111i\u0323nh sai<\/p>\n}

A. z_{1<\/sub>\u00a0+ z2<\/sub>\u00a0= 3 + i \u00a0\u00a0\u00a0B. z1<\/sub>\u00a0– z2<\/sub>\u00a0= 1 + 5i<\/p>\n}

C. z_{1<\/sub>.z2<\/sub> = 8 – i \u00a0\u00a0\u00a0D.z1<\/sub>. z2<\/sub>\u00a0= 8 + i<\/p>\n}

C\u00e2u 2:<\/b>\u00a0Cho hai s\u00f4\u0301 ph\u01b0\u0301c z_{1<\/sub>= – 3 + 4i, z2<\/sub>\u00a0= 4 – 3i . M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = z1<\/sub>\u00a0+ z2<\/sub>\u00a0+ z1<\/sub>. z2<\/sub>\u00a0la\u0300<\/p>\n}

A. 27\u00a0\u00a0\u00a0B. \u221a27\u00a0\u00a0\u00a0C. \u221a677\u00a0\u00a0\u00a0D. 677.<\/p>\n

C\u00e2u 3:<\/b>\u00a0Ti\u0300m ca\u0301c s\u00f4\u0301 th\u01b0\u0323c x, y sao cho: (1 – 2i)x + (1 + 2i)y = 1 + i<\/p>\n

<\/p>\n

C\u00e2u 4:<\/b>\u00a0Ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = (3 + 4i)(4 – 3i) + (2 – i)(3 + 2i) la\u0300<\/p>\n

A. 32 va\u0300 8i\u00a0\u00a0\u00a0B.32 va\u0300 8 \u00a0\u00a0\u00a0C. 18 va\u0300 -14\u00a0\u00a0\u00a0D. 32 va\u0300 -8<\/p>\n

C\u00e2u 5:<\/b>\u00a0Cho ca\u0301c s\u00f4\u0301 ph\u01b0\u0301c z_{1<\/sub>\u00a0= -1 + i, z2<\/sub>\u00a0= 1 – 2i, z3<\/sub>\u00a0= 1 + 2i . 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}

B. 1\u00a0\u00a0\u00a0B. \u221a13\u00a0\u00a0\u00a0C. 5\u00a0\u00a0\u00a0D. 13.<\/p>\n

C\u00e2u 6:<\/b><\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n

1-D<\/td>\n

2-C<\/td>\n

3-A<\/td>\n

4-B<\/td>\n

5-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 1:<\/b><\/p>\n T\u00f4\u0309ng cu\u0309a z1<\/sub>\u00a0va\u0300 z2<\/sub>\u00a0la\u0300 z1<\/sub>\u00a0+ z2<\/sub>\u00a0= (2 + 1) + (3 – 2)i = 3 + i<\/p>\n Hi\u00ea\u0323u cu\u0309a z1<\/sub>\u00a0va\u0300 z2<\/sub>\u00a0la\u0300 z1<\/sub>\u00a0– z2<\/sub>\u00a0= (2 – 1) + (3 + 2)i = 1 + 5i<\/p>\n Ti\u0301ch cu\u0309a z1<\/sub>\u00a0va\u0300 z2<\/sub>\u00a0la\u0300 z1<\/sub>. z2<\/sub>\u00a0= (2 + 3i)(1 – 2i) = 2 – 4i + 3i – 6i2<\/sup>\u00a0= 2 – i + 6 = 8 – i<\/p>\n V\u00e2\u0323y cho\u0323n \u0111a\u0301p a\u0301n D.<\/p>\n C\u00e2u 2:<\/b><\/p>\n Ta co\u0301<\/p>\n <\/p>\n Do \u0111o\u0301 z = z1<\/sub>\u00a0+ z2<\/sub>\u00a0+ z1<\/sub>. z2<\/sub>\u00a0= 1 + i + 25i = 1 + 26i<\/p>\n <\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n C.<\/p>\n C\u00e2u 3:<\/b><\/p>\n Ta co\u0301<\/p>\n (1 – 2i)x + (1 + 2i)y = 1 + i <=> (x + y) + (2y – 2x)i = 1 + i<\/p>\n <\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n A.<\/p>\n C\u00e2u 4:<\/b><\/p>\n Ta co\u0301<\/p>\n z = (12 – 9i + 16i – 12i2<\/sup>) + (6 + 4i – 3i – 2i2<\/sup>) = (12 + 7i + 12) + (6 + i + 2) = 32 + 8i<\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n B.<\/p>\n C\u00e2u 5:<\/b><\/p>\n Ta co\u0301:<\/p>\n z2<\/sub>z3<\/sub>\u00a0= (1 – 2i)(1 + 2i) = 1 – 4i2<\/sup>\u00a0= 5<\/p>\n z1<\/sub>z2<\/sub>\u00a0+ z1<\/sub>z3<\/sub>\u00a0= z1<\/sub>(z2<\/sub>\u00a0+ z3<\/sub>) = (-1 + i)(1 – 2i + 1 + 2i) = -2 + 2i<\/p>\n Suy ra<\/p>\n <\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n B.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Cho hai s\u00f4\u0301 ph\u01b0\u0301c z1\u00a0= 2 + 3i, z2\u00a0= 1 – 2i . Ti\u0300m kh\u0103\u0309ng \u0111i\u0323nh sai A. z1\u00a0+ z2\u00a0= 3 + i \u00a0\u00a0\u00a0B. z1\u00a0– z2\u00a0= 1 + 5i C. z1.z2 = 8 – i \u00a0\u00a0\u00a0D.z1. z2\u00a0= 8 + i C\u00e2u 2:\u00a0Cho hai s\u00f4\u0301 ph\u01b0\u0301c z1= – 3 + 4i, z2\u00a0= 4 – 3i […]<\/p>\n","protected":false},"author":3,"featured_media":26825,"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