a/ Bạn tự giải
b/ \(\left\{{}\begin{matrix}x-2y=0\\ax-3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2ay-3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\\left(2a-3\right)y=2\end{matrix}\right.\)
- Với \(a=\frac{3}{2}\) hệ vô nghiệm
- Với \(a\ne\frac{3}{2}\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{4}{2a-3}\\y=\frac{2}{2a-3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\) \(\Leftrightarrow2a-3< 0\Leftrightarrow a< \frac{3}{2}\)