Решебник по алгебре 8 класс Макарычев ФГОС Задание 1330

\[\boxed{\text{1330.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} 2x^{2} - xy = y^{2} + 5 \\ x^{2} - xy = y^{2} + 1\ \ \\ \end{matrix} \right.\ ( - )\]

\[x^{2} = 4\]

\[x = \pm 2.\]

\[1)\ x = 2:\]

\[x^{2} - xy = y^{2} + 1\]

\[4 - 2y = y^{2} + 1\]

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

\[y_{1} + y_{2} = - 2;\ \ y_{1} \cdot y_{2} = - 3\]

\[y_{1} = - 3;\ \ y_{2} = 1.\]

\[2)\ x = - 2:\]

\[x^{2} - xy = y^{2} + 1\]

\[4 + 2y = y^{2} + 1\]

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

\[D_{1} = 1 + 3 = 4\]

\[y_{1} = 1 + 2 = 3;\ \ \]

\[y_{2} = 1 - 2 = - 1.\]

\[Ответ:(2; - 3);(2;1);( - 2;3);\]

\[( - 2; - 1).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 3x^{2} - 2y^{2} = 2xy - 1 \\ 2x^{2} - y^{2} = 2xy - 1\ \ \\ \end{matrix} \right.\ ( - )\]

\[x^{2} - y^{2} = 0\]

\[(x - y)(x + y) = 0\]

\[x = y;\ \ x = - y.\]

\[2)\ x = y:\]

\[2x^{2} - y^{2} = 2xy - 1\]

\[2y^{2} - y^{2} = 2y^{2} - 1\]

\[y^{2} = 1\]

\[y = \pm 1.\]

\[x = \pm 1.\]

\[2)\ x = - y:\]

\[2x^{2} - y^{2} = 2xy - 1\]

\[2y^{2} - y^{2} = - 2y^{2} - 1\]

\[3y^{2} = - 1\]

\[нет\ решений.\]

\[Ответ:( - 1; - 1);\ \ (1;1).\]