Ta có: \(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(x+1+2x-2\right)\left[\left(x+1\right)-\left(2x-2\right)\right]=0\)
\(\Leftrightarrow\left(3-x\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{3;\frac{1}{3}\right\}\)