\(a,m=2\Leftrightarrow\left\{{}\begin{matrix}2x+2y=-3\\y-x=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{3}{4}\\ b,HPT\Leftrightarrow\left\{{}\begin{matrix}mx+mx\left(m-1\right)=-3\left(1\right)\\y=x\left(m-1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow mx+m^2x-mx=-3\\ \Leftrightarrow m^2x=-3\\ \Leftrightarrow x=-\dfrac{3}{m^2}\\ \Leftrightarrow y=-\dfrac{3\left(m-1\right)}{m^2}=\dfrac{3\left(1-m\right)}{m^2}\)
Để \(\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{m^2}< 0\left(luôn.đúng\right)\\\dfrac{3\left(1-m\right)}{m^2}< 0\end{matrix}\right.\)
\(\Leftrightarrow3\left(1-m\right)< 0\left(m^2>0\right)\\ \Leftrightarrow1-m< 0\left(3>0\right)\\ \Leftrightarrow m>1\)
