\({\left( {x-7} \right)^2} < \sqrt {11} \left( {x-7} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {x-7} \right)^2}-\sqrt {11} \left( {x-7} \right) < 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {x-7} \right)\left( {x-7-\sqrt {11} } \right) < 0.\)
Решим полученное неравенство методом интервалов. Найдём нули:
\(\left( {x-7} \right)\left( {x-7-\sqrt {11} } \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x-7 = 0,\\x-7-\sqrt {11} = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 7,\\x = 7 + \sqrt {11} .\end{array} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {7;7 + \sqrt {11} } \right).\)
Ответ: \(\left( {7;7 + \sqrt {11} } \right).\)