Ta có pt \(x^2-2x-1=2\left(x-1\right)\sqrt{x^2+2x-1}\)
Đặt \(\sqrt{x^2+2x-1}=a\left(a\ge0\right)\), ta có pt
\(\Leftrightarrow a^2-4x=2\left(x-1\right)a\Leftrightarrow a^2-2\left(x-1\right)a-4x=0\)
Ta có \(\Delta_a=\left(x-1\right)^2+4x=\left(x+1\right)^2\Rightarrow\left[{}\begin{matrix}a=\dfrac{2\left(x-1\right)-x-1}{2}=\dfrac{x-3}{2}\\a=\dfrac{2\left(x-1\right)+x+1}{2}=\dfrac{3x-1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2a=x-3\\2a=3x-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x^2+2x-1}=x-3\\2\sqrt{x^2+2x-1}=3x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=x^2-6x+9\\4\left(x^2+2x-1\right)=9x^2-6x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^2+14x-13=0\\5x^2-14x+5=0\end{matrix}\right.\) ...
