ĐKXĐ: \(x\ne\left\{-1;0\right\}\)
Đặt \(\left(\frac{2x}{x+1}\right)^2=t>0\) ta được:
\(t+\frac{1}{t}=2\Leftrightarrow t^2-2t+1=0\Leftrightarrow t=1\)
\(\Leftrightarrow\left(\frac{2x}{x+1}\right)^2=1\Leftrightarrow4x^2=x^2+2x+1\)
\(\Leftrightarrow3x^2-2x-1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)