Bất phương trình $\left( {{x}^{3}}-9\text{x} \right)\ln \left( x+5...

Điều kiện: x > -5
Xét dấu hàm số $f\left( x \right)=x\left( x-3 \right)\left( x+3 \right).$

$\Rightarrow \left\{ \begin{aligned}
& f\left( x \right)\ge 0\Leftrightarrow x\in \left[ -3;0 \right]\cup \left[ 3;+\infty \right) \\
& f\left( x \right)\le 0\Leftrightarrow x\in \left( -\infty ;-3 \right]\cup \left[ 0;3 \right) \\
\end{aligned} \right.$
$\left( {{x}^{3}}-9\text{x} \right)\ln \left( x+5 \right)\le 0\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {{x}^{3}}-9\text{x}\ge 0 \\
& \ln \left( x+5 \right)\le 0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& {{x}^{3}}-9\text{x}\le 0 \\
& \ln \left( x+5 \right)\ge 0 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& x\left( x-3 \right)\left( x+3 \right)\ge 0 \\
& \ln \left( x+5 \right)\le {{e}^{0}} \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& x\left( x-3 \right)\left( x+3 \right)\le 0 \\
& \ln \left( x+5 \right)\ge {{e}^{0}} \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& x\in \left[ -3;0 \right]\cup \left[ 3;+\infty \right) \\
& x\le -4 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& x\in \left( -\infty ;-3 \right]\cup \left[ 0;3 \right] \\
& x\ge -4 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& -4\le x\le -3 \\
& 0\le x\le 3 \\
\end{aligned} \right.$
Lại có $x\in \mathbb{Z}\Rightarrow x=\left\{ -4;-3;0;1;2;3 \right\}$