Ta co:
\(\Delta'=\left(m-1\right)^2-1\left(m-3\right)=m^2-2m+1-m+3=m^2-3m+4=\left(m^2-1,5\right)^2+1,75>0\)
nên pt co 2 nghiệm phân biệt
b, theo hệ thưc vi-et ta co:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)=2m-2\\x_1x_2=m-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x_1^2+2x_1x_2+x_2^2=4m^2-8m+4\\2x_1x_2=2m-6\end{matrix}\right.\Rightarrow x_1^2+x_2^2=4m^2-10m+10=\left(2m-\frac{5}{2}\right)^2+3,75\ge3,75.\text{Xay ra}\Leftrightarrow m=\frac{5}{4}\)