a: Để hệ có nghiệm duy nhất thì \(\dfrac{2m}{8}\ne\dfrac{1}{m}\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
Để hệ có vô số nghiệm thì \(\dfrac{2m}{8}=\dfrac{1}{m}=\dfrac{2}{m+2}\)
=>\(\left\{{}\begin{matrix}m^2=4\\2m=m+2\end{matrix}\right.\Leftrightarrow m=2\)
Để hệ vô nghiệm thì \(\dfrac{2m}{8}=\dfrac{1}{m}\ne\dfrac{2}{m+2}\)
=>\(\left\{{}\begin{matrix}m^2=4\\2m\ne m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m\ne2\end{matrix}\right.\)
=>m=-2
b: \(\left\{{}\begin{matrix}2mx+y=2\\8x+my=m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2mx+2\\8x+m\left(-2mx+2\right)=m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x-2m^2x=m+2-2m\\y=-2mx+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(8-2m^2\right)=2-m\\y=-2mx+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{m-2}{2\left(m-2\right)\left(m+2\right)}=\dfrac{1}{2\left(m+2\right)}\\y=-\dfrac{2m}{2\left(m+2\right)}+2=-\dfrac{m}{m+2}+2=\dfrac{-m+2m+4}{m+2}=\dfrac{m+4}{m+2}\end{matrix}\right.\)
\(-4x+y=\dfrac{-4}{2\left(m+2\right)}+\dfrac{m+4}{m+2}=\dfrac{-2}{m+2}+\dfrac{m+4}{m+2}\)
\(=\dfrac{m+4-2}{m+2}=\dfrac{m+2}{m+2}=1\)
=>-4x+y là hệ thức không phụ thuộc vào m
