Решебник по алгебре 8 класс Макарычев ФГОС Задание 1304

\[\boxed{\text{1304.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[Пусть\ \text{x\ }и\ y - скорости\ \]

\[работы\ слесарей.\ \]

\[Примем\ всю\ работу\ за\ 1.\]

\[Составим\ систему\ уравнений:\]

\[\left\{ \begin{matrix} \frac{1}{x} - 7 = \frac{1}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{1}{2(x + y)} + \frac{1}{2y} = \frac{1}{x + y} \neq \frac{9}{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} y = \frac{1}{\frac{1}{x} - 7\ }\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{1}{2(x + y)} + \frac{1}{2} \cdot \left( \frac{1}{x} - 7 \right) = \frac{1}{x + y} \neq \frac{9}{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = \frac{x}{1 - 7x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{1}{2(x + y)} - \frac{1}{x + y} = \frac{16}{2} - \frac{1}{2x} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = \frac{x}{1 - 7x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ - \frac{1}{2(x + y)} = 8 - \frac{1}{2x} \\ \end{matrix} \right.\ \]

\[- \frac{1}{2\left( x + \frac{x}{1 - 7x} \right)} =\]

\[= 8 - \frac{1}{2x}\ \ \ \ \ | \cdot 2\]

\[- \frac{1 \cdot (1 - 7x)}{x(1 - 7x) + x} = 16 - \frac{1}{x}\]

\[\frac{7x - 1}{2x - 7x^{2}} = \frac{16x - 1}{x}\]

\[(7x - 1)x =\]

\[= (16x - 1)(2x - 7x^{2})\]

\[7x^{2} - x = 32x^{2} - 112x^{3} -\]

\[- 2x + 7x^{2}\ \ \ |\ :x\]

\[7x - 1 = 32x - 112x^{2} - 2 + 7x\]

\[112x^{2} - 32x + 1 = 0\]

\[x_{1,2} = \frac{32 \pm 24}{224}\]

\[x_{1} = \frac{1}{4},\ \ x_{2} = \frac{1}{28}\]

\[y_{1} - отрицательным\ не\ \]

\[может\ быть.\]

\[y_{2} = \frac{1}{21}\]

\[t_{1} = 28\ ч;\ \ t_{2} = 21\ ч - время\ \]

\[слесарей.\]

\[Ответ:28\ ч\ и\ 21\ ч.\]