Với \(m=0\Rightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\) (thỏa mãn)
Với \(m\ne0\) hệ pt tương đương:
\(\left\{{}\begin{matrix}mx+y=2\\x-my=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2x+my=2m\\x-my=2\end{matrix}\right.\) \(\Rightarrow\left(m^2+1\right)x=2m+2\)
\(\Rightarrow x=\frac{2m+2}{m^2+1}\Rightarrow y=2-mx=\frac{2-2m}{m^2+1}\)
\(\left\{{}\begin{matrix}x\ge0\\y\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{2m+2}{m^2+1}\ge0\\\frac{2-2m}{m^2+1}\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m+1\ge0\\1-m\ge0\end{matrix}\right.\) \(\Leftrightarrow-1\le m\le1\)
