Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{-2}\)
=>\(m^2\ne-2\)(luôn đúng)
vậy: hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}x+my=2\\mx-2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2-my\\m\left(2-my\right)-2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2-my\\2m-m^2y-2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2-my\\y\left(m^2+2\right)=2m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{2m-1}{m^2+2}\\x=2-\dfrac{2m^2-m}{m^2+2}=\dfrac{2m^2+4-2m^2+m}{m^2+2}=\dfrac{m+4}{m^2+2}\end{matrix}\right.\)
x>0; y<0
=>\(\left\{{}\begin{matrix}m+4>0\\2m-1< 0\end{matrix}\right.\)
=>\(-4< m< \dfrac{1}{2}\)
