Найдите область определения функции y=корень из(x^2+2x-80)/(3x-36).

\[y = \frac{\sqrt{x^{2} + 2x - 80}}{3x - 36}\]

\[x^{2} + 2x - 80 \geq 0\]

\[D = 1 + 80 = 81\]

\[x_{1} = 1 + 9 = 10;\ \ \ \]

\[x_{2} = 1 - 9 = - 8\]

\[(x + 8)(x - 10) \geq 0.\]

\[3x - 36 > 0\]

\[3x > 36\]

\[x > 12.\]

\[Ответ:x \in ( - \infty; - 10\rbrack \cup \lbrack 8;12) \cup (12; + \infty).\]