\(\left\{{}\begin{matrix}4x-my=m-4\\\left(2m+6\right)x+y=2m+1\end{matrix}\right.\)
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{4}{2m+6}< >\dfrac{-m}{1}\)
=>\(-2m^2-6m< >4\)
=>\(-2m^2-6m-4\ne0\)
=>\(-2\left(m^2+3m+2\right)\ne0\)
=>\(m^2+3m+2\ne0\)
=>\(\left(m+1\right)\left(m+2\right)\ne0\)
=>\(\left\{{}\begin{matrix}m+1\ne0\\m+2\ne0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\ne-1\\m\ne-2\end{matrix}\right.\)
=>\(m\notin\left\{-1;-2\right\}\)
Để hệ phương trình vô nghiệm thì \(\dfrac{4}{2m+6}=\dfrac{-m}{1}\ne\dfrac{m-4}{2m+1}\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{2m+6}=-m\\\dfrac{-m}{1}\ne\dfrac{m-4}{2m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2m^2-6m=4\\-2m^2-m\ne m-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2m^2-6m-4=0\\-2m^2-2m+4\ne0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2+3m+2=0\\m^2+m-2\ne0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m+1\right)\left(m+2\right)=0\\\left(m+2\right)\left(m-1\right)\ne0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m+1=0\\m+2=0\end{matrix}\right.\\\left\{{}\begin{matrix}m+2\ne0\\m-1\ne0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{-1;-2\right\}\\m\notin\left\{-2;1\right\}\end{matrix}\right.\Leftrightarrow m=-1\)
Để hệ phương trình có vô số nghiệm thì \(\dfrac{4}{2m+6}=\dfrac{-m}{1}=\dfrac{m-4}{2m+1}\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{2m+6}=-m\\\dfrac{m-4}{2m+1}=-m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-m=\dfrac{2}{m+3}\\m-4=-m\left(2m+1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-m^2-3m=2\\m-4+2m^2+m=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2+3m=-2\\2m^2+2m-4=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2+3m+2=0\\m^2+m-2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m+2\right)\left(m+1\right)=0\\\left(m+2\right)\left(m-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{-2;-1\right\}\\m\in\left\{-2;1\right\}\end{matrix}\right.\)
=>m=-2