C\u00e2u 28:<\/b>\u00a0M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = -1 + 7i la\u0300<\/p>\n
A. 7\u00a0\u00a0\u00a0B. 6\u00a0\u00a0\u00a0C. \u221a50\u00a0\u00a0\u00a0D. 8<\/p>\n
C\u00e2u 29:<\/b>\u00a0C\u0103n b\u00e2\u0323c hai cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = -8 + 6i la\u0300<\/p>\n
A. -1 – 3i va\u0300 1 + 3i \u00a0\u00a0\u00a0B. -1 + 3i va\u0300 1 – 3i<\/p>\n
C. 3 + i va\u0300 -3 – i \u00a0\u00a0\u00a0D. -3 + i va\u0300 -3 – i.\u221a2<\/p>\n
C\u00e2u 30:<\/b>\u00a0Tr\u00ean t\u00e2\u0323p s\u00f4\u0301 ph\u01b0\u0301c, ph\u01b0\u01a1ng tri\u0300nh x2<\/sup>\u00a0+ 2x + 3 = 0 co\u0301 nghi\u00ea\u0323m la\u0300<\/p>\n A. 1 – \u221a2i va\u0300 1 + \u221a2i \u00a0\u00a0\u00a0 B. -1 – \u221a2i va\u0300 -1 + \u221a2i<\/p>\n C. 1 + \u221a2i va\u0300 -1 + \u221a2i \u00a0\u00a0\u00a0D. 1 + \u221a2i va\u0300 -1 – \u221a2i<\/p>\n C\u00e2u 31:<\/b>\u00a0Ph\u01b0\u01a1ng tri\u0300nh z2<\/sup>\u00a0+ 4z + 7 co\u0301 hai nghi\u00ea\u0323m z1<\/sub>, z2<\/sub>\u00a0. Gia\u0301 tri\u0323 cu\u0309a bi\u00ea\u0309u th\u01b0\u0301c T = |z1<\/sub>|2<\/sup>\u00a0+ |z2<\/sub>|2<\/sup>\u00a0b\u0103\u0300ng<\/p>\n A. 7\u00a0\u00a0\u00a0 B. 2\u221a7 \u00a0\u00a0\u00a0C. 14\u00a0\u00a0\u00a0 D. 25<\/p>\n C\u00e2u 32:<\/b>\u00a0Cho ca\u0301c s\u00f4\u0301 ph\u01b0\u0301c z1<\/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<\/p>\n <\/p>\n A. 1 \u00a0\u00a0\u00a0B. 3 \u00a0\u00a0\u00a0C. 4 \u00a0\u00a0\u00a0D. 5.<\/p>\n C\u00e2u 33:<\/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’ = (z + i)(z\u2212<\/i>\u00a0+ i) la\u0300 m\u00f4\u0323t s\u00f4\u0301 th\u01b0\u0323c va\u0300 la\u0300 \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng co\u0301 ph\u01b0\u01a1ng tri\u0300nh<\/p>\n A. x = 0 \u00a0\u00a0\u00a0B. y = 0 \u00a0\u00a0\u00a0 C. x = y \u00a0\u00a0\u00a0 D. x = -y<\/p>\n C\u00e2u 34:<\/b>\u00a0Cho s\u00f4\u0301 ph\u01b0\u0301c z co\u0301 m\u00f4\u0111un b\u0103\u0300ng 1. Gia\u0301 tri\u0323 nho\u0309 nh\u00e2\u0301t cu\u0309a bi\u00ea\u0309u th\u01b0\u0301c<\/p>\n <\/p>\n A. 2 \u00a0\u00a0\u00a0 B. 0 \u00a0\u00a0\u00a0C. -2 \u00a0\u00a0\u00a0D. -1<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 28:<\/b><\/p>\n <\/p>\n C\u00e2u 29:<\/b><\/p>\n Ta c\u00f3: z = -8 + 6i = 9i2<\/sup>\u00a0+ 6i + 1 = (3i + 1)2<\/sup><\/p>\n Do \u0111\u00f3 c\u00e1c c\u0103n b\u1eadc hai c\u1ee7a z l\u00e0 \u00b1(1 + 3i)<\/p>\n Ch\u00fa \u00fd: C\u00f3 th\u1ec3 g\u1ecdi c\u0103n b\u1eadc hai c\u1ee7a z = -8 + 6i l\u00e0 w = a + bi (a, b \u2208 R). Ta c\u00f3:<\/p>\n w2<\/sup>\u00a0= a2<\/sup>\u00a0+ 2abi + b2<\/sup>i2<\/sup>\u00a0= a2<\/sup>\u00a0– b2<\/sup>\u00a0+ 2abi<\/p>\n Ta c\u00f3:<\/p>\n <\/p>\n V\u1eady c\u00e1c c\u0103n b\u1eadc hai c\u1ee7a z = -8 + 6i l\u00e0 -1 – 3i v\u00e0 1 + 3i<\/p>\n C\u00e2u 30:<\/b><\/p>\n Ta c\u00f3: \u0394’ = 12<\/sup>\u00a0– 3 = -2 = 2i2<\/sup><\/p>\n V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 z1,2<\/sub>\u00a0= -1 \u00b1 \u221a2i<\/p>\n C\u00e2u 31:<\/b><\/p>\n Ta c\u00f3: \u0394 4 – 7 = -3 = 3i2<\/sup><\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m z1,2<\/sub>\u00a0= -2 \u00b1 i\u221a3.<\/p>\n V\u1eady T = 2(\u221a7)2<\/sup>\u00a0= 14<\/p>\n C\u00e2u 32:<\/b><\/p>\n Ta c\u00f3: z1<\/sub>z2<\/sub>\u2212<\/i>\u00a0+ z2<\/sub>z3<\/sub>\u00a0+ z3<\/sub>z1<\/sub>\u2212<\/i>\u00a0+ (1 – 2i)= (-1 + i)(1 + 2i) + (1 – 2i)(1 + 2i) + (1 + 2i)(-1 – i)<\/p>\n = (1 + 2i)( -1 + i -1 – i) + 1 – 4i2<\/sup>\u00a0= -2(1 + 2i) + 1 + 4 = 3 – 4i<\/p>\n <\/p>\n C\u00e2u 33:<\/b><\/p>\n \u0110\u1eb7t z = a + bi (a, b \u2208 R). Ta c\u00f3<\/p>\n z’ = [a + (b + 1)i][a – (b – 1)i] = a2<\/sup>\u00a0+ (b – 1)2<\/sup>\u00a0+ 2ai<\/p>\n Do \u0111\u00f3 z’ \u2208 R <=> a = 0.<\/p>\n V\u1eady t\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m M(a; b) bi\u1ec3u di\u1ec5n s\u1ed1 ph\u1ee9c z l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh z = 0<\/p>\n C\u00e2u 34:<\/b><\/p>\n \u0110\u1eb7t z = a + bi (a, b \u2208 R). Ta c\u00f3 |z| = 1 n\u00ean a2<\/sup>\u00a0+ b2<\/sup>\u00a0= 1 v\u00e0 1\/z =\u00a0z\u2212<\/i>\u00a0= a – bi<\/p>\n Suy ra T = (a + bi)2<\/sup>\u00a0+ (a – bi)2<\/sup>\u00a0= 2(a2<\/sup>\u00a0– b2<\/sup>) = 2(2a2<\/sup>\u00a0– 1) \u2265 -2<\/p>\n D\u1ea5u \u201c=\u201d x\u1ea3y ra khi a = 0. V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a T l\u00e0 -2<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 28:\u00a0M\u00f4\u0111un cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = -1 + 7i la\u0300 A. 7\u00a0\u00a0\u00a0B. 6\u00a0\u00a0\u00a0C. \u221a50\u00a0\u00a0\u00a0D. 8 C\u00e2u 29:\u00a0C\u0103n b\u00e2\u0323c hai cu\u0309a s\u00f4\u0301 ph\u01b0\u0301c z = -8 + 6i la\u0300 A. -1 – 3i va\u0300 1 + 3i \u00a0\u00a0\u00a0B. -1 + 3i va\u0300 1 – 3i C. 3 + i va\u0300 -3 – i \u00a0\u00a0\u00a0D. -3 […]<\/p>\n","protected":false},"author":3,"featured_media":26563,"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 28-C<\/td>\n 29-A<\/td>\n 30-B<\/td>\n 31-C<\/td>\n 32-D<\/td>\n 33-A<\/td>\n 34-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n