Đáp án:
$\begin{array}{l}
2)4\left( {x - 8} \right) < 0\\
\Rightarrow x - 8 < 0\\
\Rightarrow x < - 8\\
Vậy\,x < - 8\\
5)\left( {x - 3} \right)\left( { - {x^2} - 4} \right) < 0\\
\Rightarrow x - 3 > 0\\
\Rightarrow x > 3\\
Vậy\,x > 3\\
11)\left( {x + 3} \right)\left( {x - 4} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 3 > 0\\
x - 4 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 3 < 0\\
x - 4 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 3\\
x > 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 3\\
x < 4
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x > 4\\
x < - 3
\end{array} \right.\\
Vậy\,x < - 3\,hoac\,x > 4\\
14)\left( {{x^2} - 13} \right)\left( {{x^2} - 17} \right) < 0\\
\Rightarrow 13 < {x^2} < 17\\
\Rightarrow {x^2} = 16\\
\Rightarrow \left[ \begin{array}{l}
x = 4\\
x = - 4
\end{array} \right.
\end{array}$
