Решите систему неравенств: x^2-4x-12<=0; x^2-6x-7<0.

Ответ:

\[\left\{ \begin{matrix} x² - 4x - 12 \leq 0\ \ (1) \\ x² - 6x - 7 < 0\ \ \ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ x_{1} + x_{2} = 4,\ \ \ x_{1} \cdot x_{2} = - 12\]

\[x_{1} = 6,\ \ x_{2} = - 2\]

\[(2)\ x_{1} + x_{2} = 6,\ \ \ x_{1} \cdot x_{2} = - 7\]

\[x_{1} = 7,\ \ x_{2} = - 1\]

\[Ответ:( - 1;6\rbrack.\]