Найдите область определения функции y=корень из (x^2-4x-12)/(2x-18).

\[y = \frac{\sqrt{x^{2} - 4x - 12}}{2x - 18}\]

\[x^{2} - 4x - 12 \geq 0\]

\[x_{1} + x_{2} = 4;\ \ \ x_{1} \cdot x_{2} = - 12\]

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

\[(x + 2)(x - 6) \geq 0.\]

\[2x - 18 > 0\]

\[2x > 18\]

\[x > 9.\]

\[Ответ:x \in ( - \infty; - 2\rbrack \cup \lbrack 6;9) \cup (9; + \infty).\]