Решебник по алгебре 8 класс Макарычев ФГОС Задание 683

\[\boxed{\text{683\ (683).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{2} + px + q = 0,\ \ x_{1}^{2} + x_{2}^{2} - ?\]

\[\left\{ \begin{matrix} x_{1} + x_{2} = - p \\ x_{1}x_{2} = q\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[x_{1}^{2} + x_{2}^{2} =\]

\[= x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2} - 2x_{1}x_{2} =\]

\[= \left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} =\]

\[= ( - p)^{2} - 2q = p^{2} - 2q\]

\[Ответ:p^{2} - 2q.\ \]

\[\boxed{\text{683.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ xy = 2;\ \ \ \ y = \frac{2}{x}\text{.\ }\]

\[Целые\ решения:\]

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

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

\[\textbf{б)}\ x² - y^{2} = 3\]

\[(x - y)(x + y) = 3\]

\[x - y = \pm 1;\]

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

\[Целые\ решения:\]

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

\[( - 2;1).\]