Решите систему неравенств: x^2+x-12<=0; x>2.

Ответ:

\[\left\{ \begin{matrix} x² + x - 12 \leq 0 \\ x > 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + x - 12 = 0\]

\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 4\]

\[x_{1} \cdot x_{2} = - 12,\ \ x_{2} = 3\]

\[Ответ:(2;3\rbrack.\]