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

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

\[\left\{ \begin{matrix} \frac{x - 2}{x - 3} > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 7)(x + 1) \leq 0 \\ \end{matrix} \right.\ \]

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

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

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

\[Ответ:x \in \lbrack - 1;2) \cup (3;\ 7\rbrack.\]