Để hệ có nghiệm duy nhất thì \(\dfrac{2m}{8}\ne\dfrac{1}{m}\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
\(\left\{{}\begin{matrix}2mx+y=2\\8x+my=m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2m^2x+my=2m\\8x+my=m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(2m^2-8\right)=2m-m-2=m-2\\2mx+y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{m-2}{2\left(m-2\right)\left(m+2\right)}=\dfrac{1}{2\left(m+2\right)}\\y=2-2mx=2-\dfrac{2m}{2\left(m+2\right)}=2-\dfrac{m}{m+2}=\dfrac{2m+4-m}{m+2}=\dfrac{m+4}{m+2}\end{matrix}\right.\)
Để có hệ có vô số nghiệm thì \(\dfrac{2m}{8}=\dfrac{1}{m}=\dfrac{2}{m+2}\)
=>\(\dfrac{m}{4}=\dfrac{1}{m}=\dfrac{2}{m+2}\)
=>\(\left\{{}\begin{matrix}m^2=4\\m\left(m+2\right)=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\\left(m+4\right)\left(m-2\right)=0\end{matrix}\right.\)
=>m=2
Để hệ vô nghiệm thì \(\dfrac{2m}{8}=\dfrac{1}{m}\ne\dfrac{2}{m+2}\)
=>\(\left\{{}\begin{matrix}2m^2=8\\m+2\ne2m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=4\\-m\ne-2\end{matrix}\right.\)
=>m=-2
