Найдите область определения функции f(x)=(x^2-5)/(x^2-6x-16).

\[f(x) = \frac{x^{2} - 5}{x^{2} - 6x - 16}\]

\[x^{2} - 6x - 16
eq 0\]

\[x_{1} + x_{2} = 6;\ x_{1} \cdot x_{2} = - 16\]

\[x_{1}
eq - 2;\ x_{2} = 8.\]

\[D(f) = ( - \infty; - 2) \cup ( - 2;8) \cup (8; + \infty).\]