\[\sqrt{\left( x^{2} + 6x \right)^{- 1}} = \frac{1}{\sqrt{x² - 6x}}\]
\[x^{2} + 6x > 0\]
\[x(x + 6) > 0\]
\[Ответ:\]
\[при\ \ x \in ( - \infty; - 6) \cup (0; + \infty)\text{.\ }\]