biến đổi tẹo =))
\(Pt\Leftrightarrow x^2+\dfrac{x^2}{\left(x+1\right)^2}-\dfrac{2x^2}{x+1}+\dfrac{2x^2}{x+1}=0\)
\(\Leftrightarrow\left(x-\dfrac{x}{x+1}\right)^2+\dfrac{2x^2}{x+1}=3\)
\(\Leftrightarrow\dfrac{x^4}{\left(x+1\right)^2}+\dfrac{2x^2}{x+1}=3\)
đặt \(\dfrac{x^2}{x+1}=a\) => pt <=> a2+2a-3=0......
(nhớ DKXD đấy )