Ta có \(a+b+c=1-2\left(m-1\right)+2m-3=0\)
\(\Rightarrow\) phương trình luôn có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=1\\x_2=2m-3\end{matrix}\right.\)
TH1: \(x_2=x_1^2\Rightarrow2m-3=1^2\Rightarrow2m=4\Rightarrow m=2\)
TH2: \(x_1=x_2^2\Rightarrow\left(2m-3\right)^2=1\Rightarrow\left(2m-4\right)\left(2m-2\right)=0\Rightarrow\left[{}\begin{matrix}m=1\\m=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=1\\m=2\end{matrix}\right.\)