Заметим, что \(100{x^2}-140x + 49 = {\left( {10x-7} \right)^2}.\)
\(\dfrac{{{{\log }_8}\left( {{4^x}-2} \right)-{{\log }_8}{2^x}}}{{100{x^2}-140x + 49}} \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{{{\log }_8}\left( {{4^x}-2} \right)-{{\log }_8}{2^x}}}{{{{\left( {10x-7} \right)}^2}}} \le 0.\)
Тогда последнее неравенство равносильно следующей системе:
\(\left\{ \begin{array}{l}{4^x}-2 > 0,\\{2^x} > 0,\\10x-7 \ne 0,\\{\log _8}\left( {{4^x}-2} \right)-{\log _8}{2^x} \le 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}{2^{2x}} > 2,\\x \in R,\\x \ne 0,7,\\{\log _8}\left( {{4^x}-2} \right) \le {\log _8}{2^x}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}x > 0,5,\\x \ne 0,7,\\{4^x}-2 \le {2^x}.\end{array} \right.\)
Решим неравенство \({4^x}-2 \le {2^x}.\) Пусть \({2^x} = t.\) Тогда:
\({t^2}-2 \le t\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{t^2}-t-2 \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,-1 \le t \le 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,-1 \le {2^x} \le 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \le 1.\)
Следовательно,
\(\left\{ \begin{array}{l}x > 0,5,\\x \ne 0,7,\\{4^x}-2 \le {2^x}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}x > 0,5,\\x \ne 0,7,\\x \le 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x \in \left( {0,5;0,7} \right) \cup \left( {0,7;1} \right].\)
Ответ: \(\left( {0,5;0,7} \right)\, \cup \left( {0,7;1} \right].\)