Решебник по алгебре 8 класс Мерзляк Задание 612

\[\boxed{\mathbf{612\ (612).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\frac{1}{a^{2} - 4b^{2}}\ :\frac{a + 2b}{(2b - a)^{2}} =\]

\[= \frac{(2b - a)^{2}}{(a - 2b)(a + 2b)(a + 2b)} =\]

\[= \frac{a - 2b}{(a + 2b)^{2}}\]

\[2)\ \frac{a - 2b}{a(a + 2b)} - \frac{a - 2b}{(a + 2b)^{2}} =\]

\[= \frac{a^{2} - 4b^{2} - a^{2} + 2ab}{a(a + 2b)^{2}} =\]

\[= \frac{- 4b^{2} + 2ab}{a(a + 2b)^{2}}\]

\[\frac{2b}{a^{2}} = \frac{2b}{a^{2}} \Longrightarrow что\ и\ требовалось\ \]

\[доказать.\]

\[1)\ \frac{2a}{a + 3} - \frac{4a}{(a + 3)^{2}} =\]

\[= \frac{2a^{2} + 6a - 4a}{(a + 3)^{2}} = \frac{2a(a + 1)}{(a + 3)^{2}}\]

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

\[= \frac{2a(a + 1)(a - 3)(a + 3)}{(a + 3)^{2}(a + 1)} =\]

\[= \frac{2a(a - 3)}{a + 3}\]

\[3)\ \frac{2a^{2} - 6a}{a + 3} - \frac{a^{2} - 9a}{a + 3} =\]

\[= \frac{2a^{2} - 6a - a^{2} + 9a}{a + 3} =\]

\[= \frac{a(a + 3)}{(a + 3)} = a\ \ \ \]

\[a = a \Longrightarrow что\ и\ требовалось\ \]

\[доказать.\]