\(\dfrac{{-18}}{{{{\left( {x + 4} \right)}^2}-10}} \ge 0\left| { \cdot \left( {-1} \right)} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{18}}{{{{\left( {x + 4} \right)}^2}-10}} \le 0.\)
Решим полученное неравенство методом интервалов. Найдём нули знаменателя:
\({\left( {x + 4} \right)^2}-10 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {x + 4} \right)^2} = 10\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x + 4 = -\sqrt {10} ,\\x + 4 = \sqrt {10} \end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -4-\sqrt {10} ,\\x = -4 + \sqrt {10} .\end{array} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \left( {-4-\sqrt {10} ;-4 + \sqrt {10} } \right).\)
Ответ: \(\left( {-4-\sqrt {10} ;-4 + \sqrt {10} } \right).\)