Để hệ có nghiệm duy nhất thì \(\dfrac{m+1}{m^2}\ne\dfrac{-2}{-1}=2\)
=>\(2m^2\ne m+1\)
=>\(2m^2-m-1=0\)
=>\(\left(m-1\right)\left(2m+1\right)\ne0\)
=>\(m\notin\left\{1;-\dfrac{1}{2}\right\}\)
\(\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\m^2x-y=m^2+2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\2m^2x-2y=2m^2+4m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(2m^2-m-1\right)=2m^2+4m-m+1\\m^2x-y=m^2+2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2m^2+3m+1}{2m^2-m-1}=\dfrac{\left(2m+1\right)\left(m+1\right)}{\left(m-1\right)\left(2m+1\right)}=\dfrac{m+1}{m-1}\\y=m^2x-m^2-2m=\dfrac{m^2\left(m+1\right)}{m-1}-\dfrac{\left(m^2+2m\right)\left(m-1\right)}{m-1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{m+1}{m-1}\\y=\dfrac{m^3+m^2-m^3+m^2-2m^2+2m}{m-1}=\dfrac{2m}{m-1}\end{matrix}\right.\)
Để x,y đều nguyên thì \(\left\{{}\begin{matrix}m+1⋮m-1\\2m⋮m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m-1+2⋮m-1\\2m-2+2⋮m-1\end{matrix}\right.\)
=>\(2⋮m-1\)
=>\(m-1\in\left\{1;-1;2;-2\right\}\)
=>\(m\in\left\{2;0;3;-1\right\}\)
