Найдите область определения функции: y=√(x^2+2x-24)/(2x-16).

\[y = \frac{\sqrt{x^{2} + 2x - 24}}{2x - 16}\ \]

\[2x - 16
eq 0\]

\[2x
eq 16\]

\[x
eq 8.\]

\[x^{2} + 2x - 24 \geq 0\]

\[D = 1 + 24 = 25\]

\[x_{1} = - 1 + 5 = 4;\]

\[x_{2} = - 1 - 5 = - 6.\]

\[(x + 6)(x - 4) \geq 0\]

\[Но\ x
eq 8:\]

\[x \in ( - \infty; - 6\rbrack \cup \lbrack 4;8) \cup (8; + \infty).\]