Câu hỏi: Có bao nhiêu số nguyên $x$ thỏa mãn $\left[ {{\log }_{3}}\left[ {{x}^{2}}+1 \right]-{{\log }_{3}}\left[ x+21 \right] \right]\left[ 16-{{2}^{x-1}} \right]\ge 0?$
A. $17$.
B. $18$.
C. $16$.
D. Vô số.
Lời giải
Điều kiện $x>-21$.
$\begin{aligned}
& \left[ {{\log }_{3}}\left[ {{x}^{2}}+1 \right]-{{\log }_{3}}\left[ x+21 \right] \right]\left[ 16-{{2}^{x-1}} \right]\ge 0\Leftrightarrow \left\{ \begin{aligned}
& x>-21 \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& lo{{g}_{3}}\left[ {{x}^{2}}+1 \right]-{{\log }_{3}}\left[ x+21 \right]\ge 0 \\
& 16-{{2}^{x-1}}\ge 0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& lo{{g}_{3}}\left[ {{x}^{2}}+1 \right]-{{\log }_{3}}\left[ x+21 \right]\le 0 \\
& 16-{{2}^{x-1}}\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& x>-21 \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& lo{{g}_{3}}\left[ {{x}^{2}}+1 \right]\ge {{\log }_{3}}\left[ x+21 \right] \\
& 16\ge {{2}^{x-1}} \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& lo{{g}_{3}}\left[ {{x}^{2}}+1 \right]\le {{\log }_{3}}\left[ x+21 \right] \\
& 16\le {{2}^{x-1}} \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>-21 \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& \left[ {{x}^{2}}+1 \right]\ge \left[ x+21 \right] \\
& x\le 5 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& \left[ {{x}^{2}}+1 \right]\le \left[ x+21 \right] \\
& x\ge 5 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& x>-21 \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x\ge 5 \\
& x\le -4 \\
\end{aligned} \right. \\
& x\le 5 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& -4\le x\le 5 \\
& x\ge 5 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>-21\ \left[ 1 \right] \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x\ge 5 \\
& x\le -4 \\
\end{aligned} \right. \\
& x\le 5 \\
\end{aligned} \right.\ \left[ 2 \right] \\
& \left\{ \begin{aligned}
& -4\le x\le 5 \\
& x\ge 5 \\
\end{aligned} \right.\ \left[ 3 \right] \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned}$
Từ $\left[ 1 \right],\left[ 2 \right]$ ta có $\left[ \begin{aligned}
& x=5 \\
& -21