1: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{2}\ne-\dfrac{2}{-m}\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
\(\left\{{}\begin{matrix}mx-2y=2m-1\\2x-my=9-3m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2mx-4y=4m-2\\2mx-m^2y=9m-3m^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(m^2-4\right)=4m-2-9m+3m^2\\2x-my=9-3m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{3m^2-5m-2}{m^2-4}=\dfrac{\left(m-2\right)\left(3m+1\right)}{\left(m-2\right)\left(m+2\right)}=\dfrac{3m+1}{m+2}\\2x=9-3m+my\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{3m+1}{m+2}\\2x=9-3m+\dfrac{3m^2+m}{m+2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{3m+1}{m+2}\\2x=\dfrac{\left(-3m+9\right)\left(m+2\right)+3m^2+m}{m+2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{3m+1}{m+2}\\x=\dfrac{-3m^2+6m+9m+18+3m^2+m}{2m+4}=\dfrac{16m+18}{2\left(m+2\right)}=\dfrac{8m+9}{m+2}\end{matrix}\right.\)
2:
b: Để x,y đều là số nguyên thì \(\left\{{}\begin{matrix}3m+1⋮m+2\\8m+9⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3m+6-5⋮m+2\\8m+16-11⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-5⋮m+2\\-11⋮m+2\end{matrix}\right.\Leftrightarrow m+2\inƯC\left(-5;-11\right)\)
=>\(m+2\in\left\{1;-1\right\}\)
=>\(m\in\left\{-1;-3\right\}\)
