a: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{1}{-1}\)
=>\(m\ne-1\)
b: \(\left\{{}\begin{matrix}x+y=1\\mx-y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y+mx-y=1+2m\\x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(m+1\right)=2m+1\\y=1-x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2m+1}{m+1}\\y=1-\dfrac{2m+1}{m+1}=\dfrac{m+1-2m-1}{m+1}=-\dfrac{m}{m+1}\end{matrix}\right.\)
x+2y=2
=>\(\dfrac{2m+1}{m+1}+\dfrac{-2m}{m+1}=2\)
=>\(\dfrac{1}{m+1}=2\)
=>\(m+1=\dfrac{1}{2}\)
=>\(m=-\dfrac{1}{2}\left(nhận\right)\)
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