Để pt có nghiệm duy nhất
\(\dfrac{m}{4}\ne\dfrac{1}{m}\Rightarrow m^2\ne4\Leftrightarrow m\ne\pm2\)
\(\left\{{}\begin{matrix}m^2x+my=2m\\4x+my=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-4\right)x=2m-4\\y=2-mx\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{m+2}\\y=2-\dfrac{2m}{m+2}=\dfrac{m+2-2m}{m+2}=\dfrac{2-m}{m+2}\end{matrix}\right.\)
Ta có : x > 0 <=> \(\dfrac{2}{m+2}>0\Rightarrow m+2>0\Leftrightarrow m>-2\)
Với y > 0 <=> \(\dfrac{2-m}{m+2}>0\)
TH1 : \(\left\{{}\begin{matrix}2-m>0\\m+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 2\\m>-2\end{matrix}\right.\)<=> -2 < m < 2
TH2 : \(\left\{{}\begin{matrix}2-m< 0\\m+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>2\\m< -2\end{matrix}\right.\)(vô lí)