ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1303

\[\boxed{\text{1303.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[Пусть\ \ x\ и\ y - скорости\]

\(\ самосвалов.\)

\[Примем\ всю\ руду\ за\ 1.\]

\[Составим\ систему\ уравнений:\]

\[\left\{ \begin{matrix} \frac{1}{x} + 3 = \frac{1}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{1}{3x} + \frac{2}{3y} = \frac{1}{x + 3} + 7\frac{1}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{1}{x} = \frac{1}{y} - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{1}{3} \cdot \frac{1}{x} + \frac{2}{3y} = \frac{1}{x + y} + \frac{22}{3}\text{\ \ }(1) \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \frac{y}{1 - 3y}\ \ \ \ \ \ подставляем\ в\ (1) \\ \frac{1}{3} \cdot \left( \frac{1}{y} - 3 \right) + \frac{2}{3y} = \frac{1}{\frac{y}{1 - 3y} + y} + \frac{22}{3} \\ \end{matrix} \right.\ \]

\[\frac{1}{3y} - 1 + \frac{2}{3y} =\]

\[= \frac{1(1 - 3y)}{y + y(1 - 3y)} + \frac{22}{3}\]

\[\frac{3}{3y} - 1 = \frac{1 - 3y}{y + y(1 - 3y)} + \frac{22}{3}\]

\[\frac{1 - y}{y} = \frac{3 - 9y + 44y - 66y^{2}}{3 \cdot \left( 2y - 3y^{2} \right)}\]

\[\frac{1 - y}{y} = \frac{3 + 35y - 66y^{2}}{3 \cdot \left( 2y - 3y^{2} \right)}\]

\[3y + 35y^{2} - 66y^{3} =\]

\[= 3 \cdot \left( 2y - 3y^{2} \right)(1 - y)\]

\[3y + 35y^{2} - 66y^{3} = 6y -\]

\[- 9y^{2} - 6y^{2} + 9y³\]

\[75y^{3} - 50y^{2} + 3y = 0\]

\[y\left( 75y^{2} - 50y + 3 \right) = 0,\]

\[\ \ y = 0 \Longrightarrow не\ может\ быть.\]

\[75y^{2} - 50y + 3 = 0\]

\[D = 2500 - 4 \cdot 3 \cdot 75 = 1600\]

\[y_{1,2} = \frac{50 \pm 400}{150} = \frac{3}{5};\ \ \ \frac{1}{15}\]

\[x_{1} = \frac{3}{5}\ :\left( 1 - \frac{9}{5} \right) < 0 \Longrightarrow не\ \]

\[может\ быть.\]

\[x_{2} = \frac{1}{15}\ :\left( 1 - \frac{3}{15} \right) =\]

\[= \frac{1}{15} \cdot \frac{15}{12} = \frac{1}{12}\]

\[то\ есть,\ t_{1} = 15\ ч,\]

\[\text{\ \ }t_{2} = 12\ ч - время\ самосвалов.\]

\[Ответ:15\ ч\ и\ 12\ ч.\]