ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 815

\[\left( 6x^{2} + 7x + 2 \right)\left( 6x^{2} + 7x + 1 \right) = 210\]

\[y = 6x^{2} + 7x + 1:\]

\[(y + 1)y = 210\]

\[y^{2} + y - 210 = 0\]

\[D = 1 + 840 = 841\]

\[y_{1} = \frac{- 1 - 29}{2} = - 15;\ \]

\[y_{2} = \frac{- 1 + 29}{2} = 14.\]

\[1)\ y = - 15:\]

\[6x^{2} + 7x + 1 = - 15\]

\[6x^{2} + 7x + 16 = 0\]

\[D = 49 - 384 = - 335 < 0\]

\[x \in \varnothing.\]

\[2)\ y = 14:\]

\[6x^{2} + 7x + 1 = 14\]

\[6x^{2} + 7x - 13 = 0\]

\[D = 49 + 312 = 361\]

\[x_{1} = \frac{- 7 - 19}{2 \bullet 6} = - \frac{13}{6};\]

\[x_{2} = \frac{- 7 + 19}{2 \bullet 6} = 1.\]

\[Ответ:\ - \frac{13}{6};\ 1.\]

\[2)\ (x + 1)(x + 2)(x - 2)(x - 3) = 10\]

\[\left( x^{2} - x - 2 \right)\left( x^{2} - x - 6 \right) = 10\]

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

\[y(y - 4) = 10\]

\[y^{2} - 4y - 10 = 0\]

\[D = 16 + 40 = 56\]

\[x = \frac{4 \pm \sqrt{56}}{2} = \frac{4 \pm 2\sqrt{14}}{2} =\]

\[= 2 \pm \sqrt{14}.\]

\[Вернем\ замену:\]

\[x^{2} - x - 2 = 2 \pm \sqrt{14}\]

\[x^{2} - x - \left( 4 \pm \sqrt{14} \right) = 0\]

\[D = 1^{2} + 4\left( 4 \pm \sqrt{14} \right) =\]

\[= 17 \pm 4\sqrt{14}\]

\[x = \frac{1 \pm \sqrt{17 \pm 4\sqrt{14}}}{2}.\]

\[Ответ:\ \ \frac{1 \pm \sqrt{17 \pm 4\sqrt{14}}}{2}.\]