ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 382

\[\boxed{\mathbf{382}.}\]

\[6x^{4} - 11x^{3} - 13x^{2} + 10x +\]

\[+ 8 = 0;\ \ \ x_{1} = 1;\ \ x_{2} = - \frac{2}{3}.\]

\[P(1) = 6 - 11 - 13 + 10 +\]

\[+ 8 = 0;\]

\[P\left( - \frac{2}{3} \right) = 6 \cdot \frac{16}{81} + 11 \cdot \frac{8}{27} -\]

\[- 13 \cdot \frac{4}{9} - 10 \cdot \frac{2}{3} + 8 =\]

\[= \frac{96}{81} + \frac{264}{81} - \frac{468}{81} - \frac{540}{81} + 8 =\]

\[= - 8 + 8 = 0.\]

\[P(x) = (x - 1)\left( x + \frac{2}{3} \right)\]

\[\left( 6x^{2} - 9x - 12 \right)\]

\[6x^{2} - 9x - 12 = 0\ \ \ \ \ |\ :3\]

\[2x^{2} - 3x - 4 = 0\]

\[D = 9 + 32 = 41\]

\[x = \frac{3 \pm \sqrt{41}}{4}.\]

\[Ответ:\ \ x = - \frac{2}{3};\ \ 1;\ \ \frac{3 \pm \sqrt{41}}{4}.\]