ĐKXĐ: \(x\ne1\)
\(\Leftrightarrow x^2+\left(\dfrac{x}{x-1}\right)^2+2x.\dfrac{x}{x-1}-\dfrac{2x^2}{x-1}=1\)
\(\Leftrightarrow\left(x+\dfrac{x}{x-1}\right)^2-\dfrac{2x^2}{x-1}-1=0\)
\(\Leftrightarrow\left(\dfrac{x^2}{x-1}\right)^2-\dfrac{2x^2}{x-1}-1=0\)
Đặt \(\dfrac{x^2}{x-1}=t\)
\(\Rightarrow t^2-2t-1=0\)
\(\Leftrightarrow...\)