[2(x+1/x)-5][(x+1/x)+1]=0这个方程要怎么解

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[2(x+1/x)-5][(x+1/x)+1]=0这个方程要怎么解
[2(x+1/x)-5][(x+1/x)+1]=0这个方程要怎么解

[2(x+1/x)-5][(x+1/x)+1]=0这个方程要怎么解
令(x+1/x)=k,则(2k-5)(k+1)=0
x+1/x=k=2.5或-1
若x+1/x=2.5,则x=2或0.5
若x+1/x=-1,则无解
所以,x=2或1/2

令t=x+1/x (t∈(2,正无穷))
原式化为(2t-5)(t+1)=0
t=5/2或t=-1(舍)
则x+1/x=5/2
易得x=2或1/2

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