Hệ có nghiệm duy nhất \(\Leftrightarrow m\ne\pm2\)
+) Nếu \(m=0\) thì \(\left(x;y\right)=\left(4;\frac{5}{2}\right)\) T/m \(x,y>0\)
+) Nếu \(m\ne0;m\ne\pm2\) thì
\(\left\{{}\begin{matrix}x=\frac{8-m}{m+2}\\y=\frac{5}{m+2}\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}\frac{8-m}{m+2}>0\\\frac{5}{m+2}>0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\frac{8-m}{m+2}>0\\m>-2\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}8-m>0\\m>-2\end{matrix}\right.\)
\(\Leftrightarrow8>m>-2\)