Решебник по алгебре 8 класс Макарычев ФГОС Задание 686

\[\boxed{\text{686\ (686).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[3x^{2} + 2x + k = 0,\]

\[\ \ 2x_{1} = - 3x_{2} \Longrightarrow x_{1} = - 1,5x_{2}\]

\[x^{2} + \frac{2}{3}x + \frac{k}{3} = 0\]

\[Ответ:k = - 8\text{.\ }\]

\[\boxed{\text{686.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \frac{x^{2} - 1}{6x^{2}}\ :\frac{x^{2} + x}{3} =\]

\[= \frac{(x - 1)(x + 1) \cdot 3}{6x^{2} \cdot x \cdot (x + 1)} = \frac{x - 1}{2x^{3}}\]

\[\textbf{б)}\ \frac{16n^{2} - 1}{n^{2} - 2n}\ \ :\frac{8n}{3n - 6} =\]

\[= \frac{\left( 16n^{2} - 1 \right) \cdot 3 \cdot (n - 2)}{n(n - 2) \cdot 8n} =\]

\[= \frac{48n^{2} - 3}{8n^{2}}\]

\[\textbf{в)}\ \frac{x - 4}{y^{2} - xy}\ \ :\frac{5x - 20}{x^{2} - xy} =\]

\[= \frac{(x - 4) \cdot x(x - y)}{y(y - x) \cdot 5(x - 4)} = - \frac{x}{5y}\]