Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx+m^2y=m^2+m\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(m^2-1\right)=m^2-m\\x+my=m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\dfrac{m}{m+1}\\x=m+1-\dfrac{m^2}{m+1}=\dfrac{m^2+2m+1-m^2}{m+1}=\dfrac{2m+1}{m+1}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>=2\\y>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2m-2}{m+1}>=0\\\dfrac{m-m-1}{m+1}>=0\end{matrix}\right.\)
=>\(-\dfrac{1}{m+1}>=0\)
=>m+1<0
=>m<-1
