Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx+m^2y=m^2+m\\mx+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(m^2-1\right)=m^2+m-3m+1\\x+my=m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m^2-2m+1}{\left(m-1\right)\left(m+1\right)}=\dfrac{\left(m-1\right)^2}{\left(m-1\right)\cdot\left(m+1\right)}=\dfrac{m-1}{m+1}\\x=m+1-my\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=m+1-\dfrac{m^2-m}{m+1}=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\end{matrix}\right.\)
Để x,y đều là số nguyên thì \(\left\{{}\begin{matrix}m-1⋮m+1\\3m+1⋮m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m+1-2⋮m+1\\3m+3-2⋮m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2⋮m+1\\-2⋮m+1\end{matrix}\right.\)
=>\(m+1\in\left\{1;-1;2;-2\right\}\)
=>\(m\in\left\{0;-2;1;-3\right\}\)
mà \(m\notin\left\{1;-1\right\}\)
nên \(m\in\left\{0;-2;-3\right\}\)
