Số nghiệm nguyên x của bất phương trình

Ta có: \(\left( {{3^{{x^2} - 1}} - {{27}^{x + 1}}} \right)\left[ {{{\log }_3}\left( {x + 8} \right) - 2} \right] \le 0\)\( \Leftrightarrow \left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{{3^{{x^2} - 1}} - {{27}^{x + 1}} \le 0}\\{lo{g_3}\left( {x + 8} \right) - 2 \ge 0}\end{array}} \right.\\\left\{ \begin{array}{l}{3^{{x^2} - 1}} - {27^{x + 1}} \ge 0\\lo{g_3}\left( {x + 8} \right) - 2 \le 0\end{array} \right.\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{{3^{{x^2} - 1}} \le {3^{3x + 3}}}\\{log{ _3}\left( {x + 8} \right) \ge 2}\end{array}} \right.\\\left\{ \begin{array}{l}{3^{{x^2} - 1}} \ge {3^{3x + 3}} \\lo{g_3}\left( {x + 8} \right) \le 2\end{array} \right.\end{array} \right.\)\[ \Leftrightarrow \left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{{x^2} - 1 \le 3x + 3}\\{x + 8 \ge 9}\end{array}} \right.\\\left\{ \begin{array}{l}{x^2} - 1 \ge 3x + 3\\x + 8 \le 9\\x + 8 > 0\end{array} \right.\end{array} \right.\]\( \Leftrightarrow \left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{{x^2} - 3x - 4 \le 0}\\{x \ge 1}\end{array}} \right.\\\left\{ \begin{array}{l}{x^2} - 3x - 4 \ge 0\\ - 8 < x \le 1\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{l}}{ - 1 \le x \le 4}\\{x \ge 1}\end{array}} \right.\\\left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge 4\\x \le - 1\end{array} \right.\\ - 8 < x \le 1\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}1 \le x \le 4\\ - 8 < x \le - 1\end{array} \right.\).

Mà \(x \in \mathbb{Z}\) nên tập nghiệm của bất phương trình là \[S = \left\{ { - 7\,;\,\, - 6\,;\,\, \ldots ;\,\, - 1\,;\,\,1\,;\,\, \ldots ;\,\,4} \right\}.\]

Do đó, bất phương trình có 11 nghiệm nguyên. Chọn A.