Решебник по алгебре 8 класс Макарычев ФГОС Задание 644

\[\boxed{\text{644\ (644).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{2} - 5ax + 4a^{2} = 0\]

\[D = 25a^{2} - 16a^{2} = 9a^{2}\]

\[x_{1,2} = \frac{5a \pm \sqrt{9a^{2}}}{2}\]

\[x_{1} = \frac{5a + 3a}{2} = \frac{8a}{2} = 4a\]

\[x_{2} = \frac{5a - 3a}{2} = \frac{2a}{2} = a\]

\[Ответ:x_{1} = 4a;x_{2} = a.\]

\[\textbf{б)}\ 3x^{2} - 10ax + 3a^{2} = 0\]

\[D = 100a^{2} - 36a^{2} = 64a^{2}\]

\[x_{1,2} = \frac{10a \pm \sqrt{64a^{2}}}{6}\]

\[x_{1} = \frac{10a + 8a}{6} = \frac{18a}{6} = 3a\]

\[x_{2} = \frac{10a - 8a}{6} = \frac{2a}{6} = \frac{a}{3}\]

\[Ответ:x_{1} = 3a;x_{2} = \frac{a}{3}\text{.\ }\]

\[\boxed{\text{644.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

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

\[при\ x = 3 + \sqrt{5};y = 3 - \sqrt{5}:\]

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

\[= \left( 3 + \sqrt{5} - 3 + \sqrt{5} \right) =\]

\[= \left( 2\sqrt{5} \right)^{2} = 4 \cdot 5 = 20.\ \]