Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{{x^2} - 4}}{{25}}.\left( {{x^2} - 4} \right).\frac{{{x^2} - 16}}{9}.({x^2} - 10) < 0\\
\to \frac{{{{\left( {{x^2} - 4} \right)}^2}}}{{25}}.\frac{{{x^2} - 16}}{9}.({x^2} - 10) < 0\\
Do:{\left( {{x^2} - 4} \right)^2} \ge 0\forall x \in R\\
\to \frac{{{{\left( {{x^2} - 4} \right)}^2}}}{{25}} \ge 0\forall x \in R\\
Bpt \to \frac{{{x^2} - 16}}{9}.({x^2} - 10) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 16 > 0\\
{x^2} - 10 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} - 16 < 0\\
{x^2} - 10 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 4\\
x < - 4
\end{array} \right.\\
- \sqrt {10} < x < \sqrt {10}
\end{array} \right.\left( {KTM} \right)\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > \sqrt {10} \\
x < - \sqrt {10}
\end{array} \right.\\
- 4 < x < 4
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 4 < x < - \sqrt {10} \\
\sqrt {10} < x < 4
\end{array} \right.
\end{array}\)
