x+1/x=1 求x^2019+1/x^2019=?4
发布网友
发布时间：2023-11-10 22:24
共2个回答
热心网友
时间：2024-03-14 07:55
(x+1)(x²-x+1)=0,x³+1=0
x³=-1
得x是-1的3次复数根
x^2019=(x³)^673=(-1)^673=-1
所以 x^2019+1/x^2019=(-1)+1/(-1)=-2
热心网友
时间：2024-03-14 07:54
x^2 -x+1 =0
x
= (1+√3i)/2 or (1-√3i)/2
= cos(π/3) + isin(π/3) or cos(-π/3) + isin(-π/3)
case 1 : x=cos(π/3) + isin(π/3)
x^2019 + x^(-2019)
=cos(2019π/3) + isin(2019π/3) +cos(-2019π/3) + isin(-2019π/3)
=2cos(2019π/3)
=2cos(673π)
=2cosπ
=-2
case 2 : x=cos(-π/3) + isin(-π/3)
x^2019 + x^(-2019)
=cos(-2019π/3) + isin(-2019π/3) +cos(2019π/3) + isin(2019π/3)
=2cos(2019π/3)
=-2
ie
x^2019 + x^(-2019) =-2
