Có bao nhiêu số nguyên $x$ thỏa mãn $\left[ {{\log }_{2}}\left(...

Ta có
$
\left[\log _{2}\left(x^{2}+1\right)-\log _{2}(x+31)\right]\left(32-2^{x-1}\right) \geq 0
$
$\left[ \begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{x > - 31}\\
{{{\log }_2}\left( {{x^2} + 1} \right) \ge {{\log }_2}(x + 31)}\\
{32 \ge {2^{x - 1}}}
\end{array}} \right.\\
\left\{ {\begin{array}{*{20}{l}}
{x > - 31}\\
{{{\log }_2}\left( {{x^2} + 1} \right) \le {{\log }_2}(x + 31)}\\
{32 \le {2^{x - 1}}}
\end{array}} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{x > - 31}\\
{{x^2} - x - 30 \ge 0}\\
{x - 1 \le 5}
\end{array}} \right.\\
\left\{ {\begin{array}{*{20}{l}}
{x > - 31}\\
{{x^2} - x - 30 \le 0}\\
{x - 1 \ge 5}
\end{array}} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ {\begin{array}{*{20}{l}}
{x > - 31}\\
{\left[ {\begin{array}{*{20}{l}}
{x \le - 5}\\
{x \ge 6}
\end{array}} \right.}\\
{x \le 6}
\end{array}} \right.\\
\left\{ {\begin{array}{*{20}{l}}
{x > - 31}\\
{x \in [ - 5;6]}\\
{x \ge 6}
\end{array}} \right.
\end{array} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{31 < x \le - 5}\\
{x = 6}
\end{array}} \right.$
Do $x$ nguyên nên $x \in\{-30 ;-29 ;-28 ; \ldots ;-5 ; 6\}$.
Vậy có 27 giá trị nguyên của $x$ thỏa mãn bất phương trình đã cho.