Có bao nhiêu số nguyên $x$ thỏa mãn $\left[ 2-{{\log }_{3}}\left(...

+ ${{5}^{x+1}}-{{5}^{1-x}}-24=0$ $\Leftrightarrow {{5.5}^{2x}}-{{24.5}^{x}}-5=0\Leftrightarrow \left[ \begin{aligned}
& {{5}^{x}}=5 \\
& {{5}^{x}}=-\dfrac{1}{5} \left( l \right) \\
\end{aligned} \right.\Leftrightarrow x=1$.
+ $2-{{\log }_{3}}\left( {{3}^{x}}-2 \right)=0$ $\Leftrightarrow {{\log }_{3}}\left( {{3}^{x}}-2 \right)=2\Leftrightarrow x={{\log }_{3}}11$.
+ $\left[ 2-{{\log }_{3}}\left( {{3}^{x}}-2 \right) \right]\sqrt{{{5}^{x+1}}-{{5}^{1-x}}-24}\ge 0\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {{5}^{x+1}}-{{5}^{1-x}}-24=0 \\
& {{3}^{x}}-2>0 \\
\end{aligned} \right. \left( I \right) \\
& \left\{ \begin{aligned}
& {{5}^{x+1}}-{{5}^{1-x}}-24>0 \\
& 2-{{\log }_{3}}\left( {{3}^{x}}-2 \right)\ge 0 \\
\end{aligned} \right. \left( II \right) \\
\end{aligned} \right.$
+ Giải $\left( I \right):$ $\left\{ \begin{aligned}
& {{5}^{x+1}}-{{5}^{1-x}}-24=0 \\
& {{3}^{x}}-2>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{5.5}^{2x}}-{{24.5}^{x}}-5=0 \\
& {{3}^{x}}>2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=1 \\
& x>{{\log }_{3}}2 \\
\end{aligned} \right.\Leftrightarrow x=1$.
+ Giải $\left( II \right):$ $\left\{ \begin{aligned}
& {{5}^{x+1}}-{{5}^{1-x}}-24>0 \\
& 2-{{\log }_{3}}\left( {{3}^{x}}-2 \right)\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{5}^{x}}>5 \\
& {{\log }_{3}}\left( {{3}^{x}}-2 \right)\le 2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>1 \\
& {{3}^{x}}>2 \\
& {{3}^{x}}-2\le 9 \\
\end{aligned} \right.\Leftrightarrow 1<x\le {{\log }_{3}}11$.
Vậy có 2 số nguyên $x$.