Có bao nhiêu số nguyên $x$ thỏa mãn $\left[ {{\log }_{3}}\left(...

Điều kiện: $x>-31$.
Ta có $\left[ {{\log }_{3}}\left( {{x}^{2}}+1 \right)-{{\log }_{3}}\left( x+31 \right) \right]\left( 32-{{2}^{x-1}} \right)\ge 0\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+1 \right)-{{\log }_{3}}\left( x+31 \right)\ge 0 \\
& 32-{{2}^{x-1}}\ge 0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+1 \right)-{{\log }_{3}}\left( x+31 \right)\le 0 \\
& 32-{{2}^{x-1}}\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {{x}^{2}}+1\ge x+31 \\
& {{2}^{x-1}}\le 32 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& {{x}^{2}}+1\le x+31 \\
& {{2}^{x-1}}\ge 32 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {{x}^{2}}-x-30\ge 0 \\
& {{2}^{x-1}}\le {{2}^{5}} \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& {{x}^{2}}-x-30\le 0 \\
& {{2}^{x-1}}\ge {{2}^{5}} \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x\le -5 \\
& x\ge 6 \\
\end{aligned} \right. \\
& x\le 6 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& -5\le x\le 6 \\
& x\ge 6 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x\le -5 \\
& x=6 \\
\end{aligned} \right.$.
Kết hợp với điều kiện $x>-31$ ta có $\left[ \begin{aligned}
& -31<x\le -5 \\
& x=6 \\
\end{aligned} \right.$.
Vậy có 27 số nguyên $x$.