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Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
Khi \(m\notin\left\{1;-1\right\}\) thì \(\left\{{}\begin{matrix}x+my=m+1\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y-2m=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-m^2+1\right)=-m^2+m\\x=m+1-my\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m^2-m}{m^2-1}=\dfrac{m\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\dfrac{m}{m+1}\\x=m+1-\dfrac{m^2}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m}{m+1}\\x=\dfrac{\left(m+1\right)^2-m^2}{m+1}=\dfrac{2m+1}{m+1}\end{matrix}\right.\)
Để \(\left\{{}\begin{matrix}x>=2\\y>=1\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}\dfrac{2m+1}{m+1}>=2\\\dfrac{m}{m+1}>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2\left(m+1\right)}{m+1}>=0\\\dfrac{m-m-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2m-2}{m+1}>=0\\\dfrac{-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{m+1}>=0\\-\dfrac{1}{m+1}>=0\end{matrix}\right.\Leftrightarrow m+1< 0\)
=>m<-1
x=(2m+3)/(m^2+1)
y=(3m-2)/(m^2+1)
y=x-1<=> (3m-2)/(m^2+1)=(2m+3-m^2-1)/(m^2+1)
<=>m^2+m-4=0=>\(\left[\begin{matrix}m=\frac{-1-\sqrt{17}}{2}\\m=\frac{-1+\sqrt{17}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}mx+2y=18\\2x-2y=-12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(m+2\right)x=6\\y=x+6\end{matrix}\right.\)
Để hệ có nghiệm duy nhất \(\Leftrightarrow m\ne-2\)
Khi đó ta có: \(\left\{{}\begin{matrix}x=\frac{6}{m+2}\\y=\frac{6}{m+2}+6\end{matrix}\right.\)
\(2x+y=9\Leftrightarrow\frac{12}{m+2}+\frac{6}{m+2}+6=9\)
\(\Leftrightarrow\frac{18}{m+2}=3\Rightarrow m+2=6\Rightarrow m=4\)