Có bao nhiêu số nguyên $x$ thỏa mãn...

TXĐ: $D=\left( -\infty ;-4 \right)\cup \left( 4;+\infty \right).$
Ta có:
$\begin{aligned}
& \text{lo}{{\text{g}}_{3}}\dfrac{{{x}^{2}}-16}{343}<\text{lo}{{\text{g}}_{7}}\dfrac{{{x}^{2}}-16}{27} \\
& \Leftrightarrow \text{lo}{{\text{g}}_{3}}7.\left[ \text{lo}{{\text{g}}_{7}}\left( {{x}^{2}}-16 \right)-3 \right]<\text{lo}{{\text{g}}_{7}}\left( {{x}^{2}}-16 \right)-3\text{lo}{{\text{g}}_{7}}3 \\
& \Leftrightarrow \left( \text{lo}{{\text{g}}_{3}}7-1 \right)\text{.lo}{{\text{g}}_{7}}\left( {{x}^{2}}-16 \right)<3\text{lo}{{\text{g}}_{3}}7-3\text{lo}{{\text{g}}_{7}}3 \\
& \Leftrightarrow {{\log }_{7}}\left( {{x}^{2}}-16 \right)<\dfrac{3\left( {{\log }_{3}}7-{{\log }_{7}}3 \right)}{{{\log }_{3}}7-1} \\
& \Leftrightarrow {{\log }_{7}}\left( {{x}^{2}}-16 \right)<3\left( 1+{{\log }_{7}}3 \right) \\
& \Leftrightarrow {{\log }_{7}}\left( {{x}^{2}}-16 \right)<{{\log }_{7}}{{21}^{3}} \\
& \Leftrightarrow {{x}^{2}}-16<{{21}^{3}} \\
& \Leftrightarrow -\sqrt{9277}<x<\sqrt{9277} \\
\end{aligned}$
Kết hợp điều kiện ta có $x\in \left\{ -96;-95;...;-5;5;...;95;96 \right\}$. Vậy có 184 số nguyên x thỏa mãn.