Решебник по алгебре 8 класс Мерзляк Задание 792

\[\boxed{\mathbf{792\ (792).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{3x + 2}{x^{2} + 2x + 4} + \frac{x^{2} + 39}{x^{3} - 8} =\]

\[= \frac{5}{x - 2}\]

\[- x^{2} - 14x + 15 = 0\]

\[x^{2} + 14x - 15 = 0\]

\[x_{1} + x_{2} = - 14,\ \ x_{1}x_{2} = - 15,\ \]

\[\ x_{1} = - 15,\ \ x_{2} = 1\]

\[Ответ:\ x = - 15;x = 1.\]

\[2)\ \frac{x}{x - 1} + \frac{x + 1}{x + 3} = \frac{8}{x^{2} + 2x - 3}\]

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

\[x_{1} + x_{2} = - 2,\ \ x_{1}x_{2} = - 3,\]

\[\ x_{1} = - 3,\ \ x_{2} = 1\]

\[x^{2} + 3x + x^{2} - 1 - 8 = 0\]

\[2x^{2} + 3x - 9 = 0\]

\[D = 9 + 72 = 81\]

\[x = \frac{- 3 - 9}{4} = - 3\]

\[x = \frac{- 3 + 9}{4} = 1,5\]

\[Ответ:x = 1,5.\]