Thay m=0 vào pt ta đc
\(x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
ĐỂ pt có 2 ngo pb thì \(\Delta'>0\Leftrightarrow\left(m-1\right)^2-m^2>0\Leftrightarrow-2m+1>0\Leftrightarrow m< \frac{1}{2}\)