$\left\{\begin{array}{l} 2y-x=m+1\\2x-y=m-2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 4y-2x=2m+2\\2x-y=m-2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 4y-2x+2x-y=2m+2+m-2\\2x-y=m-2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 3y=3m\\x=\dfrac{y+m-2}{2}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} y=m\\x=m-1\end{array} \right.$
$a)m=1\Leftrightarrow$ Hệ có nghiệm $\left\{\begin{array}{l} y=1\\x=0\end{array} \right.$
$b)x^2+y^2=(m-1)^2+m^2\\ =2m^2-2m+1\\ =\left(\sqrt{2}m\right)^2-2.\sqrt{2}m.\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(\sqrt{2}m-\dfrac{1}{\sqrt{2}}\right)^2+\dfrac{1}{2} \ge \dfrac{1}{2} \, \forall \, m$
Dấu "=" xảy ra $\Leftrightarrow\sqrt{2}m-\dfrac{1}{\sqrt{2}}=0\Leftrightarrow m=\dfrac{1}{2}$
