a: =>mx=1-2y và 3x+(m+1)y=-1
=>x=-2/m*y+1/m và 3*(y*-2/m+1/m)+(m+1)y=-1
=>\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{m}\cdot y+\dfrac{1}{m}\\-\dfrac{6}{m}y+\dfrac{3}{m}+\left(m+1\right)y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\left(-\dfrac{6}{m}+m+1\right)+=-1-\dfrac{3}{m}\\x=-\dfrac{2}{m}\cdot y+\dfrac{1}{m}\end{matrix}\right.\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}y\cdot\dfrac{m^2+m-6}{m}=\dfrac{-m-3}{m}\\x=-\dfrac{2}{m}\cdot y+\dfrac{1}{m}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\cdot\dfrac{\left(m+3\right)\left(m-2\right)}{m}=\dfrac{-\left(m+3\right)}{m}\\x=-\dfrac{2}{m}\cdot y+\dfrac{1}{m}\end{matrix}\right.\)
b: Nếu m=0 thì hệ vô nghiệm
Nếu m=-3 thì hệ có vô số nghiệm
Nếu m=2 thì hệ vô nghiệm
nếu m<>0; m<>-3; m<>2 thì hệ có nghiệm duy nhất là
\(\left\{{}\begin{matrix}y=\dfrac{-\left(m+3\right)}{m}:\dfrac{\left(m+3\right)\left(m-2\right)}{m}=-\dfrac{1}{m-2}\\x=\dfrac{2}{m}\cdot\dfrac{1}{m-2}+\dfrac{1}{m}=\dfrac{2+m-2}{m\left(m-2\right)}=\dfrac{m}{m\left(m-2\right)}=\dfrac{1}{m-2}\end{matrix}\right.\)
