a: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{1}\ne\dfrac{m+1}{-1}\)
=>\(m\ne-m-1\)
=>\(2m\ne-1\)
=>\(m\ne-\dfrac{1}{2}\)
b: \(\left\{{}\begin{matrix}mx+\left(m+1\right)y=1\\x-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx+\left(m+1\right)y=1\\x=y+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\left(y+2\right)+\left(m+1\right)y=1\\x=y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}my+2m+y\left(m+1\right)=1\\x=y+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\left(2m+1\right)=1-2m\\x=y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{1-2m}{2m+1}\\x=\dfrac{1-2m}{2m+1}+2=\dfrac{-2m+1+4m+2}{2m+1}=\dfrac{2m+3}{2m+1}\end{matrix}\right.\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}2m+3⋮2m+1\\-2m+1⋮2m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+1+2⋮2m+1\\-2m-1+2⋮2m+1\end{matrix}\right.\Leftrightarrow2⋮2m+1\)
mà 2m+1 lẻ(do m nguyên)
nên \(2m+1\in\left\{1;-1\right\}\)
=>\(m\in\left\{0;-1\right\}\)
