Đáp án:
$\begin{array}{l}
c)\left( {{x^2} - 13} \right)\left( {{x^2} - 17} \right) < 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 13 > 0\\
{x^2} - 17 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} - 13 < 0\\
{x^2} - 17 > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} > 13\\
{x^2} < 17
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} < 13\\
{x^2} > 17
\end{array} \right.
\end{array} \right.\\
\Rightarrow 13 < {x^2} < 17\\
\Rightarrow {x^2} = 16\\
\Rightarrow \left\{ \begin{array}{l}
x = 4\\
x = - 4
\end{array} \right.\\
Vậy\,x = 4;x = - 4\\
d)\left( {{x^2} + 1} \right)\left( {{x^2} - 49} \right) = 0\\
\Rightarrow {x^2} - 49 = 0\\
\Rightarrow {x^2} = 49\\
\Rightarrow \left[ \begin{array}{l}
x = 7\\
x = - 7
\end{array} \right.\\
Vậy\,x = - 7;x = 7
\end{array}$
