Решебник по алгебре 9 класс Мерзляк Задание 605

\[\boxed{\mathbf{605\ (605).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[x^{4} - y^{4} = x^{3} + y^{3}\]

\[\left( x^{2} - y^{2} \right)\left( x^{2} + y^{2} \right) =\]

\[= (x + y)\left( x^{2} - xy + y^{2} \right)\]

\[(x - y)(x + y)\left( x^{2} + y^{2} \right) -\]

\[- (x + y)\left( x^{2} - xy + y^{2} \right) = 0\]

\[(x + y)((x - y) \cdot\]

\[\cdot \left( x^{2} + y^{2} - \left( x^{2} - xy + y^{2} \right) \right) = 0\]

\[x = - y,\ \ то\ есть\ \text{\ x\ }и\ y\]

\[\ противоположные\ числа.\]

\[(x - y)\left( x^{2} + y^{2} \right) =\]

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

\[1)\ \left\{ \begin{matrix} x - y = x^{2} - xy + y^{2} \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x - y = 1 - xy \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + xy - y - 1 = 0 \\ x² + y² = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x(1 + y) - (y + 1) = 0 \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (y + 1)(x - 1) = 0 \\ x^{2} + y^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\ \left\{ \begin{matrix} y = - 1 \\ x = 0\ \ \ \ \\ \end{matrix} \right.\ ;\ \ \ \ \left\{ \begin{matrix} y = 0 \\ x = 1 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x - y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} = x^{2} - xy + y^{2} \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 1 + y \\ xy = 0\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} y = - 1 \\ x = 0\ \ \ \\ \end{matrix} \right.\ ;\ \ \left\{ \begin{matrix} y = 0 \\ x = 1 \\ \end{matrix} \right.\ \]

\[Общие\ решения\ систем\ \ \ \]

\[(0;\ - 1);\ (1;0).\]

\[Ответ:таких\ \text{\ x\ }\ и\ y\ не\]

\[\ существует.\]