ГДЗ по алгебре 9 класс Макарычев Задание 1052

\[\boxed{\text{1052\ (1052).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} \left( x^{2} + y^{2} \right)(x - y) = 447 \\ \text{xy}(x - y) = 210\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x³ - y^{3} = 657\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 210 \cdot \left( \frac{x}{y} + 1 + \frac{y}{x} \right) = 657. \\ \end{matrix} \right.\ \]

\[Пусть\ \frac{x}{y} = c,\]

\[210 \cdot \left( c + 1 + \frac{1}{c} \right) = 657\]

\[210c^{2} + 210c + 210 - 657c = 0\]

\[210c^{2} - 447c + 210 = 0\]

\[70c^{2} - 149c + 70 = 0\]

\[D = 22\ 201 - 19600 = 2601,\]

\[c_{1} = \frac{149 + 51}{140} = \frac{10}{7},\ \ \]

\[c_{2} = \frac{149 - 51}{140} = \frac{98}{140} = \frac{7}{10},\]

\[\Longrightarrow \frac{x}{y} = \frac{10}{7}\text{\ \ \ }или\ \ \frac{x}{y} = \frac{7}{10}.\]

\[Ответ:(10;7);( - 7;\ - 10).\]

\[Пусть\ \ \ \frac{x}{y} = c,\ \]

\[c - 1 + \frac{1}{c} = \frac{7}{6}\]

\[6c^{2} - 6c + 6 - 7c = 0\]

\[6c^{2} - 13c + 6 = 0\]

\[D = 169 - 144 = 25,\]

\[c_{1} = \frac{13 + 5}{12} = \frac{18}{12} = \frac{3}{2},\ \ \]

\[c_{2} = \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3},\]

\[\Longrightarrow \frac{x}{y} = \frac{3}{2}\text{\ \ \ }или\ \ \frac{x}{y} = \frac{2}{3}.\]

\[Ответ:(2;3);\ \ \ (3;2).\]