$ \left\{\begin{array}{l} mx-y=1\\x+my=2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx-y=1\\mx+m^2y=2m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx-y=1\\(m^2+1)y=2m-1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{y+1}{m}\\y=\dfrac{2m-1}{m^2+1}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{m+2}{m^2+1}\\y=\dfrac{2m-1}{m^2+1}\end{array} \right.$
$m$ có nghiệm $(x;y)$ nằm trong góc phần tư thứ $IV $
$\Rightarrow \left\{\begin{array}{l} x=\dfrac{m+2}{m^2+1}>0\\y=\dfrac{2m-1}{m^2+1}<0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m+2>0\\2m-1<0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m>-2\\m<\dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow -2<m<\dfrac{1}{2}$
