Решите систему неравенств: x-2>=5/(x+2); x-1<=6/x.

\[\left\{ \begin{matrix} x - 2^{\backslash x + 2} \geq \frac{5}{x + 2} \\ x - 1^{\backslash x} \leq \frac{6}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x - 2)(x + 2) - 5}{x + 2} \geq 0 \\ \frac{(x - 1)x - 6}{x} \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{2} - 4 - 5}{x + 2} \geq 0 \\ \frac{x^{2} - x - 6}{x} \leq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{x^{2} - 9}{x + 2} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{(x + 2)(x - 3)}{x} \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x - 3)(x + 3)}{x + 2} \geq 0 \\ \frac{(x + 2)(x - 3)}{x} \leq 0 \\ \end{matrix} \right.\ \]

\[Ответ:\lbrack - 3;\ - 2) \cup \left\{ 3 \right\}.\]