$\left\{\begin{array}{l} mx+y=1\\x+my=m+1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx+y=1\\mx+m^2y=m^2+m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx+y=1\\mx+m^2y-mx-y=m^2+m-1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} mx+y=1\\(m^2-1)y=\left(m-\dfrac{1-\sqrt{5}}{2}\right)\left(m-\dfrac{1+\sqrt{5}}{2}\right)\end{array} \right.$
$\circledast m=\pm 1,$ Phương trình vô nghiệm
$\circledast m\ne\pm 1$
Hệ có nghiệm duy nhất
$ \left\{\begin{array}{l} x=\dfrac{1-y}{m}\\y=\dfrac{m^2+m-1}{m^2-1}\end{array}\right.\\ \Leftrightarrow \left\{\begin{array}{l} x=-\dfrac{1}{m^2-1}\\y=\dfrac{m^2+m-1}{m^2-1}\end{array}\right.$