Số nghiệm nguyên của bất phương trình $\left(...

Ta có: $\left( {{3}^{{{x}^{2}}-1}}-{{27}^{x+1}} \right)\left( {{\log }_{3}}\left( x+8 \right)-2 \right)\le 0$
$\Leftrightarrow \left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-1}}-{{27}^{x+1}}\ge 0 \\
& {{\log }_{3}}\left( x+8 \right)-2\le 0 \\
\end{aligned} \right.\vee \left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-1}}-{{27}^{x+1}}\le 0 \\
& {{\log }_{3}}\left( x+8 \right)-2\ge 0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-1}}\ge {{3}^{3x+3}} \\
& {{\log }_{3}}\left( x+8 \right)\le 2 \\
\end{aligned} \right.\vee \left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-1}}\le {{3}^{3x+3}} \\
& {{\log }_{3}}\left( x+8 \right)\ge 2 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-1\ge 3x+3 \\
& x+8\le 9 \\
& x+8>0 \\
\end{aligned} \right.\vee \left\{ \begin{aligned}
& {{x}^{2}}-1\le 3x+3 \\
& x+8\ge 9 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-3x-4\ge 0 \\
& x\le 1 \\
& x>-8 \\
\end{aligned} \right.\vee \left\{ \begin{aligned}
& {{x}^{2}}-3x-4\le 0 \\
& x\ge 1 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x\le -1\vee x\ge 4 \\
& -8<x\le 1 \\
\end{aligned} \right.\vee \left\{ \begin{aligned}
& -1\le x\le 4 \\
& x\ge 1 \\
\end{aligned} \right.$
$\Leftrightarrow -8<x\le -1\vee 1\le x\le 4$
Mà $x\in \mathbb{Z}$
Nên $S=\left\{ -7;-6;...;-1;1;2;3;4 \right\}$
Bất phương trình có 11 nghiệm nguyên.