Có bao nhiêu số nguyên $x$ thỏa mãn $\log _3\left(2 x^2-4...

Điều kiện: ${{x}^{2}}-2x>0\Leftrightarrow \left( -\infty ;0 \right)\cup \left( 2;+\infty \right)$
$\begin{aligned}
& {{\log }_{3}}\left( 2{{x}^{2}}-4x \right)>{{\log }_{2}}\dfrac{{{x}^{2}}-2x}{2023} \\
& \Leftrightarrow {{\log }_{3}}\left( {{x}^{2}}-2x \right)+{{\log }_{3}}2>{{\log }_{2}}\left( {{x}^{2}}-2x \right)-{{\log }_{2}}2023 \\
& \Leftrightarrow {{\log }_{3}}2.{{\log }_{2}}\left( {{x}^{2}}-2x \right)+{{\log }_{3}}2>{{\log }_{2}}\left( {{x}^{2}}-2x \right)-{{\log }_{2}}2023 \\
& \Leftrightarrow {{\log }_{2}}\left( {{x}^{2}}-2x \right).\left( 1-{{\log }_{3}}2 \right)<{{\log }_{2}}2023+{{\log }_{3}}2 \\
& \Leftrightarrow {{\log }_{2}}\left( {{x}^{2}}-2x \right)<\dfrac{{{\log }_{2}}2023+{{\log }_{3}}2}{1-{{\log }_{3}}2} \\
\end{aligned}$
Đặt $A=\dfrac{{{\log }_{2}}2023+{{\log }_{3}}2}{1-{{\log }_{3}}2}$
$\begin{aligned}
& \Leftrightarrow {{\log }_{2}}\left( {{x}^{2}}-2x \right)<A \\
& \Leftrightarrow {{x}^{2}}-2x<{{2}^{A}} \\
& \Leftrightarrow x\in \left( -54464,6;54466,7 \right) \\
\end{aligned}$
Kết hợp với điều kiện có $108928$ giá trị.