Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-y\left(m^2-1\right)=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\y\left(m^2-1\right)=m^2+m-3m+1=m^2-2m+1=\left(m-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{\left(m-1\right)^2}{\left(m-1\right)\left(m+1\right)}=\dfrac{m-1}{m+1}\\x=m+1-my=m+1-\dfrac{m^2-m}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\end{matrix}\right.\)
x+y=5
=>\(3m+1+m-1=5\left(m+1\right)\)
=>5m+5=4m
=>m=-5(nhận)
