Решите систему неравенств: (x-7)/(x+2)<=0; x^2-4x-5>=0.

\[\left\{ \begin{matrix} \frac{x - 7}{x + 2} \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4x - 5 \geq 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 4x - 5 = x^{2} - 5x + x - 5 =\]

\[= x(x - 5) + (x - 5) =\]

\[= (x - 5)(x + 1)\]

\[\left\{ \begin{matrix} \frac{x - 7}{x + 2} \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 5)(x + 1) \geq 0 \\ \end{matrix} \right.\ \]

\[Ответ:x \in ( - 2;\ - 1\rbrack \cup \lbrack 5;7\rbrack.\]