Решите систему неравенств: (x-3)/(x+8)<=0; x^2+6x-7>=0.

\[\left\{ \begin{matrix} \frac{x - 3}{x + 8} \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 6x - 7 \geq 0 \\ \end{matrix} \right.\ \]

\[x^{2} + 6x - 7 = x^{2} + 7x - x - 7 =\]

\[= x(x + 7) - (x + 7) =\]

\[= (x + 7)(x - 1)\]

\[\left\{ \begin{matrix} \frac{x - 3}{x + 8} \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 7)(x - 1) \geq 0 \\ \end{matrix} \right.\ \]

\[Ответ:x \in ( - 8;\ - 7\rbrack \cup \lbrack 1;2\rbrack\]