\(x^2+2x-1-m^2=0\Leftrightarrow\left(x-1\right)^2=m^2\)
\(\Leftrightarrow x-1=\sqrt{m^2}=\left|m\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=m\\x-1=-m\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+m\\x=1-m\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x_1=1+m\\x_2=1-m\end{matrix}\right.\)