C\u00e2u 1:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z = 2 \u2013 2i. Ti\u0300m kh\u0103\u0309ng \u0111i\u0323nh sai.<\/p>\n
A. Ph\u00e2\u0300n th\u01b0\u0323c cu\u0309a z la\u0300: 2.<\/p>\n
B. Ph\u00e2\u0300n a\u0309o cu\u0309a z la\u0300: -2.<\/p>\n
C. S\u00f4\u0301 ph\u01b0\u0301c li\u00ean h\u01a1\u0323p cu\u0309a z la\u0300\u00a0z\u2212<\/i>\u00a0= -2 + 2i.<\/p>\n
D. M\u00f4\u0111un cu\u0309a z la\u0300<\/p>\n
<\/p>\n
C\u00e2u 2:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z = -1 + 3i. Ph\u00e2\u0300n th\u01b0\u0323c, ph\u00e2\u0300n a\u0309o cu\u0309a\u00a0z\u2212<\/i>\u00a0la\u0300<\/p>\n
A. -1 va\u0300 3 \u00a0\u00a0\u00a0B. -1 va\u0300 -3 \u00a0\u00a0\u00a0C. 1 va\u0300 -3 \u00a0\u00a0\u00a0D. -1 va\u0300 -3i.<\/p>\n
C\u00e2u 3:<\/b>\u00a0M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z tho\u0309a ma\u0303n\u00a0z\u2212<\/i>\u00a0= 8 – 6i la\u0300<\/p>\n
A. 2\u00a0\u00a0\u00a0 B. 10 \u00a0\u00a0\u00a0C. 14 \u00a0\u00a0\u00a0 D. 2\u221a7<\/p>\n
C\u00e2u 4:<\/b>\u00a0Ti\u0300m ca\u0301c s\u00f4\u0301 th\u01b0\u0323c x, y sao cho (x \u2013 2y) + (x + y + 4)I = (2x + y) + 2yi.<\/p>\n
A. x = 3, y = 1 \u00a0\u00a0\u00a0 B. x = 3, y = -1<\/p>\n
C. x = -3, y = -1 \u00a0\u00a0\u00a0 D. x = -3, y = 1<\/p>\n
C\u00e2u 5:<\/b>\u00a0Hai s\u00f4\u0301 ph\u01b0\u0301c z1<\/sub>\u00a0= x – 2i, z2<\/sup>2 + yi (x, y \u2208 R) la\u0300 li\u00ean h\u01a1\u0323p cu\u0309a nhau khi<\/p>\n A. x = 2, y = -2 \u00a0\u00a0\u00a0B. x = -2, y = -2 \u00a0\u00a0\u00a0C. x = 2, y = 2 \u00a0\u00a0\u00a0 D. x = -2, y = 2<\/p>\n C\u00e2u 6:<\/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\u0300a ma\u0303n |z| = |1 + i| la\u0300<\/p>\n A. Hai \u0111i\u00ea\u0309m \u00a0\u00a0\u00a0 B. Hai \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng<\/p>\n C. \u0110\u01b0\u01a1\u0300ng tro\u0300n ba\u0301n ki\u0301nh R=2 \u00a0\u00a0\u00a0D. \u0110\u01b0\u01a1\u0300ng tro\u0300n ba\u0301n ki\u0301nh R= \u221a2 .<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b><\/p>\n S\u00f4\u0301 ph\u01b0\u0301c li\u00ean h\u01a1\u0323p cu\u0309a z la\u0300\u00a0z\u2212<\/i>\u00a0= -2 + 2i n\u00ean cho\u0323n \u0111a\u0301p a\u0301n C.<\/p>\n C\u00e2u 2:<\/b><\/p>\n Ta co\u0301 z = -1 + 3i =>\u00a0z\u2212<\/i>\u00a0= -1 – 3i<\/p>\n V\u00e2\u0323y ph\u00e2\u0300n th\u01b0\u0323c va\u0300 ph\u00e2\u0300n a\u0309o cu\u0309a\u00a0z\u2212<\/i>\u00a0la\u0300 -1 va\u0300 -3.<\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n B.<\/p>\n C\u00e2u 3:<\/b><\/p>\n Ta co\u0301<\/p>\n <\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n B.<\/p>\n C\u00e2u 4:<\/b><\/p>\n Ta co\u0301 (x \u2013 2y) + (x + y + 4)I = (2x + y) + 2yi.<\/p>\n <\/p>\n V\u00e2\u0323y x = -3, y = 1.<\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n D.<\/p>\n C\u00e2u 5:<\/b><\/p>\n Ta co\u0301\u00a0z1<\/sub>\u2212<\/i>\u00a0= x + 2i.<\/p>\n Do \u0111o\u0301, hai s\u00f4\u0301 ph\u01b0\u0301c \u0111a\u0303 cho go\u0323i la\u0300 li\u00ean h\u01a1\u0323p cu\u0309a nhau khi va\u0300 chi\u0309 khi<\/p>\n <\/p>\n V\u00e2\u0323y x= 2, y = 2. Cho\u0323n \u0111a\u0301p a\u0301n C.<\/p>\n C\u00e2u 6:<\/b><\/p>\n Ta co\u0301 |1 + i| = \u221a(1 + 1) = \u221a2. Go\u0323i M la\u0300 \u0111i\u00ea\u0309m bi\u00ea\u0309u di\u00ea\u0303n cu\u0309a z ta co\u0301 |z| = OM.<\/p>\n Do \u0111o\u0301: |z| = |1 + i| <=> OM = \u221a2<\/p>\n V\u00e2\u0323y t\u00e2\u0323p h\u01a1\u0323p ca\u0301c \u0111i\u00ea\u0309m M bi\u00ea\u0309u di\u00ea\u0303n s\u00f4\u0301 ph\u01b0\u0301c z la\u0300 \u0111\u01b0\u01a1\u0300ng tro\u0300n t\u00e2m O, ba\u0301n ki\u0301nh R= \u221a2 .<\/p>\n Cho\u0323n \u0111a\u0301p a\u0301n D.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z = 2 \u2013 2i. Ti\u0300m kh\u0103\u0309ng \u0111i\u0323nh sai. A. Ph\u00e2\u0300n th\u01b0\u0323c cu\u0309a z la\u0300: 2. B. Ph\u00e2\u0300n a\u0309o cu\u0309a z la\u0300: -2. C. S\u00f4\u0301 ph\u01b0\u0301c li\u00ean h\u01a1\u0323p cu\u0309a z la\u0300\u00a0z\u2212\u00a0= -2 + 2i. D. M\u00f4\u0111un cu\u0309a z la\u0300 C\u00e2u 2:\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z = -1 + 3i. Ph\u00e2\u0300n th\u01b0\u0323c, ph\u00e2\u0300n […]<\/p>\n","protected":false},"author":3,"featured_media":26841,"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-C<\/td>\n 2-B<\/td>\n 3-B<\/td>\n 4-D<\/td>\n 5-C<\/td>\n 6-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n