Bất phương trình \(\left( {{x^3} - 9x} \right)\ln...

C

- Hướng dẫn giải

Phương pháp giải:

Giải bất phương trình  \({\log _a}x < b \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a > 1\\x < {a^b}\end{array} \right.\\\left\{ \begin{array}{l}0 < a < 1\\x > {a^b}\end{array} \right.\end{array} \right..\)

Giải chi tiết:

Điều kiện: \(x >  - 5.\) 

Xét dấu hàm số \(f\left( x \right) = x\left( {x - 3} \right)\left( {x + 3} \right)\)

\( \Rightarrow \left\{ \begin{array}{l}f\left( x \right) \ge 0 \Leftrightarrow x \in \left[ { - 3;\,0} \right] \cup \left[ {3; + 8} \right)\\f\left( x \right) \le 0 \Leftrightarrow x \in \left( { - \infty ; - 3} \right] \cup \left[ {0;\,3} \right]\end{array} \right.\)

\(\begin{array}{l}\left( {{x^3} - 9x} \right)\ln \left( {x + 5} \right) \le 0 \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{x^3} - 9x \ge 0\\\ln \left( {x + 5} \right) \le 0\end{array} \right.\\\left\{ \begin{array}{l}{x^3} - 9x \le 0\\\ln \left( {x + 5} \right) \ge 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x\left( {x - 3} \right)\left( {x + 3} \right) \ge 0\\x + 5 \le {e^0}\end{array} \right.\\\left\{ \begin{array}{l}x\left( {x - 3} \right)\left( {x + 3} \right) \le 0\\x + 5 \ge {e^0}\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \in \left[ { - 3;\,0} \right] \cup \left[ {3; + 8} \right)\\x \le  - 4\end{array} \right.\\\left\{ \begin{array}{l}x \in \left( { - \infty ; - 3} \right] \cup \left[ {0;\,3} \right]\\x \ge  - 4\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 4 \le x \le  - 3\\0 \le x \le 3\end{array} \right..\end{array}\)

Lại có \(x \in \mathbb{Z} \Rightarrow x \in \left\{ { - 4;\, - 3;\,0;\,\,1;\,2;\,3} \right\}\)

Chọn C.