Để hệ có nghiệm duy nhất thì \(\dfrac{m-1}{2}\ne\dfrac{-m}{-1}=m\)
=>\(2m\ne m-1\)
=>\(m\ne-1\)
\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\\left(m-1\right)x-m\left(2x-m-5\right)=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\\left(m-1\right)x-2mx+m^2+5m=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(-m-1\right)=-m^2-2m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x-m-5\\x=m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2\left(m+1\right)-m-5=2m+2-m-5=m-3\\x=m+1\end{matrix}\right.\)
\(P=x^2+y^2\)
\(=\left(m+1\right)^2+\left(m-3\right)^2\)
\(=2m^2-4m+10\)
\(=2m^2-4m+2+8\)
\(=2\left(m-1\right)^2+8>=8\forall m\)
Dấu '=' xảy ra khi m=1(nhận)
