ĐKXĐ: x<>1; x<>-1
Ta có: \(\frac{x^2-2}{x^2-1}+\frac{x}{2x-2}=\frac{2}{x+1}\)
=>\(\frac{x^2-2}{\left(x-1\right)\left(x+1\right)}+\frac{x}{2\left(x-1\right)}=\frac{2}{x+1}\)
=>\(\frac{2\left(x^2-2\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{2\cdot2\cdot\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)
=>\(2x^2-4+x^2+x=4\left(x-1\right)=4x-4\)
=>\(3x^2+x-4-4x+4=0\)
=>\(3x^2-3x=0\)
=>3x(x-1)=0
=>x(x-1)=0
=>\(\left[\begin{array}{l}x=0\left(nhận\right)\\ x=1\left(loại\right)\end{array}\right.\)
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