若z^2+z+1=0,求(1+z)*(1+z^2)*(1+z^4)*(1+z^8)…(1+z^1 若z^2+z+1=0,求(1+z)*(1+z^2)*(1+z^4)*(1+z^8 )…(1+z^1024)的值.
∵z²;+z+1=0,∴z=(-1±根號3)/2,∴z≠1.
(z-1)(z²;+z+1)=0,即z³;=1.
∴原式=(1-z)•;(1+z)•;(1+z²;)•;(1+z∧4)•;•;•;(1+z∧1024)/1—z
=(1-z∧2048)/1-z
=1-z²;╱1-z=1+z=(1±根號3)/2.
方程組X+Y=2,Y+Z=4,Z+X=8的解為幾
X=3
Y=-1
Z=5
x+y=4式1 y+Z=6式2 z+x=8式3
式1+2+3得:2x+2y+2z=18
∴x+y+z=9
∴x=9-6=3;y=9-8=1;z=9-4=5.