a/ \(\left(x+2\right)\left(x+5\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\)

+) Với \(\left\{{}\begin{matrix}x+2>0\\x+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x>-5\end{matrix}\right.\) \(\Leftrightarrow\) \(x>-2\)

+) Với \(\left\{{}\begin{matrix}x+2< 0\\x+5< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< -2\\x< -5\end{matrix}\right.\) \(\Leftrightarrow x< -2\)

Vậy =

b/ \(\left(x-2\right)\left(x+5\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2< 0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2>0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\)

+) Với \(\left\{{}\begin{matrix}x-2< 0\\x+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-5\end{matrix}\right.\) \(\Leftrightarrow-5< x< 2\)

+) Với \(\left\{{}\begin{matrix}x-2>0\\x+5< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -5\end{matrix}\right.\) \(\Leftrightarrow x\in\varnothing\)

Vậy ..