Có bao nhiêu số nguyên x thỏa mãn $\left(...

• Trường hợp 1:
$\left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-2}}-{{9}^{x+3}}\ge 0 \\
& {{\log }_{5}}\left( 2x-4 \right)-625\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-2}}\ge {{3}^{2x+6}} \\
& 0<2x-4\le 4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-2x-8\ge 0 \\
& 2<x\le 4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x\ge 4 \\
& x\le -2 \\
\end{aligned} \right. \\
& 2<x\le 4 \\
\end{aligned} \right.\Leftrightarrow x=4$
• Trường hợp 2:
$\left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-2}}-{{9}^{x+3}}\le 0 \\
& {{\log }_{5}}\left( 2x-4 \right)-625\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{3}^{{{x}^{2}}-2}}\ge {{3}^{2x+6}} \\
& 2x-4\ge 4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-2x-8\le 0 \\
& x\ge 4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -2\le x\le 4 \\
& x\ge 4 \\
\end{aligned} \right.\Leftrightarrow x=4$
Vậy có 1 giá trị nguyên.