Найдите область определения функции y=корень из (x^2+6x+8)/(3x+18).

\[y = \frac{\sqrt{x^{2} + 6x + 8}}{3x + 18}\]

\[x^{2} + 6x + 8 \geq 0\]

\[x_{1} + x_{2} = - 6;\ \ \ x_{1} \cdot x_{2} = 8\]

\[x_{1} = - 4;\ \ \ x + 2 = - 2\]

\[(x + 4)(x + 2) \geq 0.\]

\[3x + 18 > 0\]

\[3x > - 18\]

\[x > - 6.\]

\[Ответ:x \in ( - \infty; - 6) \cup ( - 6;4\rbrack \cup \lbrack - 2; + \infty).\]