Решебник по алгебре 9 класс Макарычев Задание 1053

\[\boxed{\text{1053\ (1053).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\left\{ \begin{matrix} x³ - y^{3} = 19 \cdot (x - y) \\ x³ + y³ = 7 \cdot (x + y)\text{\ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} (x - y)\left( x^{2} + xy + y^{2} \right) = 19 \cdot (x - y) \\ (x + y)\left( x^{2} - xy + y^{2} \right) = 7 \cdot (x + y)\ \\ \end{matrix} \right.\ \]

\[4)\ \left\{ \begin{matrix} x^{2} + xy + y^{2} = 19 \\ x^{2} - xy + y^{2} = 7\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2x^{2} + 2y^{2} = 19 + 7 \\ 2xy = 19 - 7\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + y^{2} = 13 \\ 2xy = 12\ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x + y)^{2} = 25 \\ xy = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} x + y = 5 \\ xy = 6\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ y = \pm 3, \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x + y = - 5 \\ xy = 6\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 3 \\ y = \pm 2. \\ \end{matrix} \right.\ \]