Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 328

\[\boxed{\mathbf{328}.}\]

\[1)\ \left( 2x^{2} - x - 1 \right)\left( 2x^{2} - x - 5 \right) -\]

\[- 5 = 0\]

\[t = 2x^{2} - x:\]

\[(t - 1)(t - 5) - 5 = 0\]

\[t^{2} - 6t + 5 - 5 = 0\]

\[t^{2} - 6t = 0\]

\[t(t - 6) = 0\]

\[t = 0;\ \ \ t = 6.\]

\[t = 0:\]

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

\[x(2x - 1) = 0\]

\[x = 0;\ \ x = 0,5.\]

\[t = 6:\]

\[2x^{2} - x = 6\]

\[2x^{2} - x - 6 = 0\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 + 7}{4} = 2;\ \]

\[\ x_{2} = \frac{1 - 7}{4} = - 1,5.\]

\[Ответ:x = - 1,5;0;0,5;2.\]

\[2)\ \left( 3x^{2} - x - 4 \right)\left( 3x^{2} - x + 2 \right) -\]

\[- 7 = 0\]

\[t = 3x^{2} - x:\]

\[(t - 4)(t + 2) - 7 = 0\]

\[t^{2} - 2t - 8 - 7 = 0\]

\[t^{2} - 2t - 15 = 0\]

\[D_{1} = 1 + 15 = 16\]

\[t_{1} = 1 + 4 = 5;\ \ \]

\[t_{2} = 1 - 4 = - 3.\]

\[t = 5:\]

\[3x^{2} - x = 5\]

\[3x^{2} - x - 5 = 0\]

\[D = 1 + 60 = 61\]

\[x = \frac{1 \pm \sqrt{61}}{6}.\]

\[t = - 3:\]

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

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

\[D = 1 - 36 = - 35 < 0\]

\[нет\ корней.\]

\[Ответ:x = \frac{1 \pm \sqrt{61}}{6}.\]