Решите систему неравенств: x-3>=7/(x+3); x-1<=12/x.

\[\left\{ \begin{matrix} x - 3^{\backslash x + 3} \geq \frac{7}{x + 3} \\ x - 1^{\backslash x} \leq \frac{12}{x}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x - 3)(x + 3) - 7}{x + 3} \geq 0 \\ \frac{(x - 1)x - 12}{x} \leq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{2} - 9 - 7}{x + 3} \geq 0\ \ \\ \frac{x^{2} - x - 12}{x} \leq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{x^{2} - 16}{x + 3} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{(x + 3)(x - 4)}{x} \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{(x - 4)(x + 4)}{x + 3} \geq 0 \\ \frac{(x + 3)(x - 4)}{x} \leq 0 \\ \end{matrix} \right.\ \]

\[Ответ:\lbrack - 4; - 3) \cup \left\{ 4 \right\}.\]