Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)(1)
\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mx+m^2y=m^2+m\\mx+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx+m^2y-mx-y=m^2+m-3m+1\\x+my=m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(m^2-1\right)=\left(m-1\right)^2\\x=m+1-my\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\left(m-1\right)^2}{\left(m+1\right)\left(m-1\right)}=\dfrac{m-1}{m+1}\\y=m+1-\dfrac{m^2-m}{m+1}=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\end{matrix}\right.\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}m-1⋮m+1\\3m+1⋮m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1-2⋮m+1\\3m+3-2⋮m+1\end{matrix}\right.\)
=>\(-2⋮m+1\)
=>\(m+1\in\left\{1;-1;2;-2\right\}\)
=>\(m\in\left\{0;-2;1;-3\right\}\)
Kết hợp với (1), ta được: \(m\in\left\{0;-2;-3\right\}\)
`{(x+my=m+1),(mx+y=3m-1):}`
`<=>{(x=m+1-my),(m(m+1-my)+y=3m-1):}`
`<=>{(x=m+1-my),(m^2+m-m^2 y+y=3m-1):}`
`<=>{(x=m+1-my),(y=[2m-1-m^2]/[1-m^2]=-[1-m]/[1+m]):}` `(m ne+-1)`
`<=>{(x=m+1+[m(1-m)]/[1+m]),(y=[m-1]/[m+1]):}`
`<=>{(x=[3m+1]/[m+1]=3-2/[m+1]),(y=[m-1]/[m+1]=1-2/[m+1]):}`
Để `x,y` đều là số nguyên `=>{(3-2/[m+1] in Z),(1-2/[m+1] in Z):}`
`=>2/[m+1] in Z`
`=>m+1 in Ư_2`
Lại có: `Ư_2 =`{`+-1;+-2`}
Mà `m in Z, m ne +-1`
`=>m=`{`-3;-2;0`}
