\[\boxed{\text{410\ (410).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
\[Решение\ квадратных\ \]
\[неравенств.\]
Решение.
\[1) - 3x^{2} + 6x + 1 > - \frac{4}{3}\text{\ \ }| \cdot 3\]
\[- 9x^{2} + 18x + 3 + 4 > 0\]
\[- 9x^{2} + 18x + 7 > 0\]
\[D = 324 + 252 = 576\]
\[x_{1,2} = \frac{- 18 \pm 24}{- 18}\]
\[x = - \frac{1}{3};\ \ \ \ x = 2\frac{1}{3}.\]
\[Ответ:x \in \left( - \frac{1}{3};2\frac{1}{3} \right)\text{.\ }\]
\[2) - 5x^{2} + 11x + 2 \leq - \frac{2}{5}\text{\ \ }| \cdot 5\]
\[- 25x^{2} + 55x + 10 + 2 \leq 0\]
\[- 25x^{2} + 55x + 12 \leq 0\]
\[D = 3025 + 1200 = 4225\]
\[x_{1,2} = \frac{- 55 \pm 65}{- 50}\]
\[x = - 0,2;\ \ \ x = 2,4\ \]
\[Ответ:x \in ( - \infty;\ - 0,2\rbrack \cup \lbrack 2,4; + \infty).\]