若x^2-x+1=0,求x^100+x^-100之值

来源：百度知道 编辑：UC知道 时间：2024/06/03 13:04:51
关键是我不知道这个代数式的变形怎么变

x^2-X+1=0
x-1+1/X=0
x+1/x=1
这个是关键后面的自己算

x^2-x+1=0
两端同时乘以 x+1得
x^3+1=0
所以 x^3=-1 所以 x^6=1 x^102=(x^6)^17=1

所以
x^100+x^(-100)
= x^100+x^(-100)*x^102
=x^96*x^3*x+x^2
=-x+x^2
=-1

x^2-x+1=0,x=cos(π/3)±isin(π/3)
当x=cos(π/3)+isin(π/3)
x^100=cos(4π/3)+isin(4π/3)
x^-100=cos(-4π/3)+isin(-4π/3)
x^100+x^-100=2cos(4π/3)=-1

当x=cos(-π/3)+isin(-π/3)
x^100=cos(-4π/3)+isin(-4π/3)
x^-100=cos(4π/3)+isin(4π/3)
x^100+x^-100=2cos(4π/3)=-1

楼上还有用三角代换的
我就不献丑了
阿米豆腐....