$\left\{\begin{matrix}(m-1)x-my=3m-1\\ 2x-y=m+5\end{matrix}\right.\\ \Leftrightarrow \left\{\begin{matrix}(m-1)x-m(2x-m-5)=3m-1\\ y=2x-m-5\end{matrix}\right.\\ \Leftrightarrow \left\{\begin{matrix}-(m+1)x=3m-1-m^2-5m\\ y=2x-m-5\end{matrix}\right.\\ \Leftrightarrow \left\{\begin{matrix}(m+1)x=(m+1)^2\\ y=2x-m-5\end{matrix}\right.\\ \Leftrightarrow \left\{\begin{matrix}x=m+1\\ y=2(m+1)-m-5\end{matrix}\right.\\ \Leftrightarrow \left\{\begin{matrix}x=m+1\\ y=m-3\end{matrix}\right.$
$P=x^2+y^2=(m+1)^2+(m-3)^2$
Theo Cô-si ta có:
$(m+1)^2+(m-3)^2\ge2\sqrt{(m+1)^2(m-3)^2}$
Dấu '=' xảy ra khi $(m+1)^2=(m-3)^2$
$\Leftrightarrow m^2+2m+1=m^2-6m+9\\ \Leftrightarrow 8m=8\Leftrightarrow m=1$
$m=1\Rightarrow P=(1+1)^2+(1-3)^2=8$
Vậy $Min_P=8$ khi $m=1$
