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

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{2}\ne\dfrac{-2}{-m}\)

=>\(m^2\ne4\)

=>\(m\notin\left\{2;-2\right\}\)

\(\left\{{}\begin{matrix}mx-2y=2m-1\\2x-my=9-3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2y=mx-2m+1\\2x-my=9-3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\2x-m\left(x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\right)=9-3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\2x-x\cdot\dfrac{m^2}{2}+m^2-\dfrac{1}{2}m=9-3m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\x\left(2-\dfrac{m^2}{2}\right)=-m^2+\dfrac{1}{2}m-3m+9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\x\cdot\dfrac{4-m^2}{2}=-m^2-\dfrac{5}{2}m+9=\dfrac{-2m^2-5m+18}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{-2m^2-5m+18}{4-m^2}=\dfrac{2m^2+5m-18}{m^2-4}\\y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{\left(2m+9\right)\left(m-2\right)}{\left(m-2\right)\left(m+2\right)}=\dfrac{2m+9}{m+2}\\y=\dfrac{2m+9}{m+2}\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2m+9}{m+2}\\y=\dfrac{2m^2+9m-2m\left(m+2\right)+m+2}{2\left(m+2\right)}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2m+9}{m+2}\\y=\dfrac{2m^2+10m+2-2m^2-4m}{2\left(m+2\right)}=\dfrac{6m+2}{2\left(m+2\right)}=\dfrac{3m+1}{m+2}\end{matrix}\right.\)

Để x,y nguyên thì \(\left\{{}\begin{matrix}2m+9⋮m+2\\3m+1⋮m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2m+4+5⋮m+2\\3m+6-5⋮m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5⋮m+2\\-5⋮m+2\end{matrix}\right.\)

=>\(5⋮m+2\)

=>\(m+2\in\left\{1;-1;5;-5\right\}\)

=>\(m\in\left\{-1;-3;3;-7\right\}\)

Đề không hiển thị hệ phương trình bạn à.

\(\left\{{}\begin{matrix}x+my=9\\mx-3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9-my\\m\left(9-my\right)-3y=4\end{matrix}\right.\)(*)

(*) <=> \(9m-m^2y-3y=4\)

<=> \(-y\left(m^2+3\right)=4-9m\) 

Vì \(m^2+3\ge3\) >0 với mọi m

=> m² + 3 khác 0

=> luôn có nghiệm y = \(\dfrac{9m-4}{m^2+3}\) với mọi m

b) Khi đó x= \(9-m.\dfrac{9m-4}{m^2+3}=\dfrac{9m^2+27-9m^2+4m}{m^2+3}=\dfrac{4m^2+27}{m^2+3}\)

Để \(x-3y=\dfrac{28}{m^2+3}-3\)

=> \(4m+27-27m+12=28-3m^2+9\)

<=> \(3m^2-3m-20m+20=0\)

<=> \(3m\left(m-1\right)-20\left(m-1\right)=0\) 

<=> \(\left(3m-20\right)\left(m-1\right)=0\)

<=> \(\left[{}\begin{matrix}m=\dfrac{20}{3}\\m=1\end{matrix}\right.\) 

Để hệ có nghiệm duy nhất thì \(\dfrac{m}{2m}\ne\dfrac{1}{3}\)

=>\(\dfrac{1}{2}\ne\dfrac{1}{3}\)(luôn đúng)

\(\left\{{}\begin{matrix}mx+y=5\\2mx+3y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2mx+2y=10\\2mx+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=4\\mx+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-4\\mx=5-y=5-\left(-4\right)=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-4\\x=\dfrac{9}{m}\end{matrix}\right.\)

\(\left(2m-1\right)\cdot x+\left(m+1\right)\cdot y=m\)

=>\(\dfrac{9}{m}\left(2m-1\right)+\left(m+1\right)\cdot\left(-4\right)=m\)

=>\(\dfrac{9\left(2m-1\right)}{m}=m+4m+4=5m+4\)

=>m(5m+4)=18m-9

=>\(5m^2-14m+9=0\)

=>(m-1)(5m-9)=0

=>\(\left[{}\begin{matrix}m=1\\m=\dfrac{9}{5}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}D=m^2-4\\D_x=9m-32\\D_y=8m-9\end{matrix}\right.\)

Hệ có nghiệm duy nhất khi \(D\ne0\Leftrightarrow m^2-4\ne0\Leftrightarrow m\ne\pm2\)

Hệ vô nghiệm khi \(\left\{{}\begin{matrix}D=0\\\left[{}\begin{matrix}D_x\ne0\\D_y\ne0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m=\pm2\\\left[{}\begin{matrix}m\ne\dfrac{32}{9}\\m\ne\dfrac{9}{8}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow m=\pm2\)

thay X=-1, Y=3 giải phương trình ta được m=3

\(m_1=\dfrac{11}{2}\)

\(m_2=3\)