Решите систему неравенств: -3x^2+16x+12<0; x^2-11x<0.

Ответ:

\[\left\{ \begin{matrix} - 3x^{2} + 16x + 12 < 0 \\ x^{2} - 11x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- 3x^{2} + 16x + 12 = 0\]

\[D = 256 + 144 = 400\]

\[x_{1} = \frac{- 16 + 20}{- 6} = - \frac{2}{3}\]

\[x_{2} = \frac{- 16 - 20}{- 6} = 6\]

\[x(x - 11) = 0\]

\[x = 0,\ \ \ \ x = 11.\]

\[Овет:x \in (6;\ \ \ 11).\]