Ta có: \(\dfrac{x+2}{x-3}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2>0\\x-3< 0\end{matrix}\right.\Leftrightarrow-2< x< 3\)
Vậy: S={x|-2<x<3}
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2< 0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2>0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -2\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x>-2\\x< 3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow-2< x< 3\)
Vậy ...