Question

已知实数x满足x2+1/x2+x+1/x=0,那么x+1/x的值为

急啊!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

x²+ 1/x²+x +1/x=0
(x+1/x)² +(x+1/x) -2=0
(x +1/x +2)(x +1/x -1)=0
x +1/x=-2或x+ 1/x=1
x+ 1/x=-2时，x²+2x+1=0 (x+1)²=0 x=-1
x+ 1/x=1时，x²-x+1=0 (x-1/2)²+3/4=0，等式左边恒>0，等式不成立，方程无解。
综上，得x=-1
x+1/x=-2

Answer 2

x2+1/x2+x+1/x=0？应该是x²+1/x²+x+1/x=0吧？

解：
x²+1/x²+x+1/x=0
x²+2+1/x²+x+1/x-2=0
x²+2×x×(1/x)+1/x²+x+1/x-2=
(x+1/x)²+(x+1/x)-2=0
令：x+1/x=u，代入上式，有：
u²+u-2=0
u²+2×(1/2)×u+(1/2)²-(1/2)²-2=0
(u+1/2)²=9/4
u+1/2=±√(9/4)
u+1/2=±3/2
u=(-1/2)±(3/2)
解得：u=1，或者：-2
即：x+1/x=1，或：x+1/x=-2