a: Để hệ có nghiệm thì \(\dfrac{2m+1}{m+3}\ne\dfrac{-3}{-\left(m+1\right)}\)
=>\(\left(2m+1\right)\left(m+1\right)\ne3\left(m+3\right)\)
=>\(2m^2+3m+1-3m-9\ne0\)
=>\(2m^2-8\ne0\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
b: \(\left\{{}\begin{matrix}\left(2m+1\right)x-3y=3m-2\\\left(m+3\right)x-\left(m+1\right)y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2m+1\right)\left(m+1\right)x-3\left(m+1\right)y=\left(3m-2\right)\left(m+1\right)\\3\left(m+3\right)x-3\left(m+1\right)y=6m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(2m^2+3m+1-3m-9\right)=\left(3m-2\right)\left(m+1\right)-6m\\\left(2m+1\right)x-3y=3m-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(2m^2-8\right)=3m^2+3m-2m-2-6m\\3y=\left(2m+1\right)x-3m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\cdot2\left(m-2\right)\left(m+2\right)=3m^2-5m-2=\left(m-2\right)\left(3m+1\right)\\3y=\left(2m+1\right)x-3m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m+1}{2\left(m+2\right)}\\3y=\left(2m+1\right)\cdot\dfrac{3m+1}{2\left(m+2\right)}-3m+2=\dfrac{6m^2+5m+1+\left(-3m+2\right)\cdot\left(2m+4\right)}{2m+4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m+1}{2\left(m+2\right)}\\3y=\dfrac{6m^2+5m+1-6m^2-12m+4m+8}{2m+4}=\dfrac{-3m+9}{2m+4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m+1}{2m+4}\\y=\dfrac{-m+3}{2m+4}\end{matrix}\right.\)
x>=2y
=>\(\dfrac{3m+1}{2m+4}-\dfrac{2\left(-m+3\right)}{2m+4}>=0\)
=>\(\dfrac{3m+1+2m-6}{2m+4}>=0\)
=>\(\dfrac{m-1}{m+2}>=0\)
=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}m>=1\\m\ne2\end{matrix}\right.\\m< -2\end{matrix}\right.\)
