Решите систему неравенств: 3x^2-8x-3>0; x<=10.

Ответ:

\[\left\{ \begin{matrix} 3x^{2} - 8x - 3 > 0 \\ x \leq 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[3x^{2} - 8x - 3 = 0\]

\[D = 64 + 36 = 100\]

\[x_{1} = \frac{8 + 10}{6} = 3\]

\[x_{2} = \frac{8 - 10}{6} = - \frac{1}{3}\]

\[Ответ:x \in \left( - \infty;\ - \frac{1}{3} \right) \cup (3;\ \ 10\rbrack.\]