2 câu trả lời
$\begin{array}{l}
\left( {x + 1} \right)\left( {x + 2} \right) > 0\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 > 0\\
x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 < 0\\
x + 2 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 1\\
x > - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 1\\
x < - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > - 1\\
x < - 2
\end{array} \right.\\
\Rightarrow S = \left( { - \infty ; - 2} \right) \cup \left( { - 1; + \infty } \right)
\end{array}$