Решебник по алгебре 9 класс Макарычев Задание 365

 

\[\boxed{\text{365\ (365).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

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

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

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

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

\[Ответ:\ - 5,6;\ \ 4.\]

\[\boxed{\text{365.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 3xy = 12\]

\[xy = 4\]

\[y = \frac{4}{x}\]

\[\textbf{б)}\frac{1}{2}xy = 6\]

\[xy = 12\]

\[y = \frac{12}{x}\]

\[\textbf{в)}\ 2xy = - 8\]

\[xy = - 4\]

\[y = - \frac{4}{x}\]

\[\textbf{г)}\ \frac{1}{2}xy = - 6\]

\[xy = - 12\]

\[y = - \frac{12}{x}\]