a: Thay x=1 và y=1 vào hệ, ta được:
\(\left\{{}\begin{matrix}-m+1=-2m\\1-m=1-m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-m+2m=-1\\0m=0\end{matrix}\right.\)
=>m=-1
b: Để hệ vô nghiệm thì \(\dfrac{-m}{1}=\dfrac{1}{-m}\ne\dfrac{-2m}{1-m}\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{1}=\dfrac{1}{m}\\\dfrac{1}{m}\ne\dfrac{2m}{1-m}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2=1\\2m^2\ne1-m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{1;-1\right\}\\2m^2+m-1< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{1;-1\right\}\\\left(m+1\right)\left(2m-1\right)< >0\end{matrix}\right.\)
=>m=1
Để hệ có vô số nghiệm thì \(\dfrac{-m}{1}=\dfrac{1}{-m}=\dfrac{-2m}{1-m}=\dfrac{2m}{m-1}\)
=>\(\left\{{}\begin{matrix}m^2=1\\-m\left(m-1\right)=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{1;-1\right\}\\-m^2+m-2m=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{1;-1\right\}\\-m^2-m=0\end{matrix}\right.\)
=>m=-1