ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 327

\[\boxed{\text{328.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\frac{1}{x + 2} - \frac{1}{x + 4} = \frac{1}{x + 8} - \frac{1}{x + 20}\]

\[\frac{x + 4 - x - 2}{(x + 2)(x + 4)} =\]

\[= \frac{x + 20 - x - 8}{(x + 8)(x + 20)}\]

\[\frac{2}{x^{2} + 6x + 8} = \frac{12}{x^{2} + 28x + 160}\]

\[x^{2} + 28x + 160 =\]

\[= 6 \cdot \left( x^{2} + 6x + 8 \right)\]

\[x^{2} + 28x + 160 =\]

\[= 6x^{2} + 36x + 48\]

\[5x^{2} + 8x - 112 = 0\]

\[D = 16 + 5 \cdot 112 = 576 = 24^{2}\]

\[x_{1,2} = \frac{- 4 \pm 24}{5} = 4;\ - 5,6.\]

\[Ответ:\ - 5,6;\ \ 4.\]