IV. TIEN TRINH DAY HOC
caub Tung do bdng 2.
y - 2 Ddp sd. Dudng thang y = - 2. Cau c. -1 o mwm 1 2 X
Ddp sd. Sd phiic thudc hinh chu nhat
cau d.
Ddp sd. Hinh trdn ban kinh 2, tam O.
Bai 6. Hudng ddn. Dua vao hai sd phiic bang nhau. cau a. Giai he phuang trinh: ^ " -^^^
[y = 2-x Ddp so. X -I, y = 1.. <-- u /-•'•,-. f2ji; + y - 1 = 0 Cau b. Giai he [x + 2y -b = 0 Ddp sd. X = - 1 , 3/ = 3.
Bai 7. Hudng ddn. Dua vao viec bieu diln hinh hgc cua sd phiic va tfnh chdt cua
tam giac vuong.
Bai 8. Hudng ddn. Dira vao phep toan vl sd phiic.
Cau a. Ta cd 2 - i + 3 - 2i = 5 - 3i. Ddpsd. 21+ i. r^' K *-' 1,1 T ' 1 + i (l + i ) ( 2 - i ) 3 + i Cau b. Cach 1. Ta co = = . 2 + i 5 5 1 + i _ ( 4 - 3 i ) ( 2 + i) + l + i _ l 2 - i Cach 2. Quy dong mdu sd : ( 4 - 3 i ) +
r,, . 23 14.
Dap so. 1.
5 5
2 +~i 2 + i 2 + i
Cau c. Cach l . T a c d (1 + i ) ^ - ( 1 - i ) ^ = ( l + i + l - i ) ( l + i - l + i) = 4 i . Cach 2. Thuc hien binh thudng : (1 + i)^ - (1 - i)^ = 2i - (-2i) = 4i
r,, . 23 14. Dap so. 1 5 5 ^ . ^ 3 + i 4 - 3 i (3 + 0 ( 2 - 0 (4 - 3 0 ( 2 + i) Lau d. — r ;:: :- = z z 2+1 2-1 5 5 Dap so. -— + —I.
5 5
Bai 9. Hudng ddn. Dua vao cac phep tinh lien hgp.
l + 8i 3 + 4i cau a. Ta cd (3 + 4i) = 2+ 5i - 1 + 3i = 1 + 8i. Tir dd ta dugc z =
Dap so.' ~ + —I.
5 5
c a u b . (4 + 7 i ) z - ( 5 - 2 0 = 6iz zz> (4 + 7 i ) z - 6 i z = 5 - 2 i = > (4 + i)z = 5 - 2 i
Dap so. 1.
Bai 10. Hudng ddn. Dua vao cdc phep tinh lien hgp. Caua. T a c d A = 49 - 96 = - 47. - 7 ± / V 4 7 Ddp so. =^ X, 2 = 2 c a u b. Ta dat u = x^ :4f.
Dap sd. Zi2 = ± ^ ; 2:3 4 = ±i<
cau c. Ta dat u = z^
Ddp sd. ± 1 va + i.
Bai 11. Hudng ddn. Rd rang (x - Zi)ix - Z2) = 0 hay X^ - iZi + Z2)X + Z1Z2 = 0
Hay x^ - 3 x + 4 = 0 .
3 + tV7 3 - iV7
^ 1 - ^ 2 — ^ 2 = — 2 —
Bai 12. Hudng ddn. Rd rang (x - Zi)ix - Z2) = 0 hay X^ - iZi + Z2)x + Z1Z2 = 0 .
HOAT DONG 3
DflP AN Bfll TflP TRAC N G H I I M