Решебник по алгебре 9 класс Макарычев Задание 914

\[\boxed{\text{914\ (914).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[= \frac{1}{2} + \frac{\left( 3m^{2} + 5m \right)(1 - 3m)}{(1 + 3m) \cdot 2 \cdot \left( 3m^{2} + 5m \right)} =\]

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

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

\[= \frac{y}{y - x} - \frac{x}{x + y} =\]

\[= \frac{xy + y² - xy + x²}{y² - x²} = \frac{y² + x²}{y² - x²}\]

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

\[= \frac{2a - 3}{2a - 3} - \frac{a - 1}{a} = 1 - \frac{a - 1}{a} =\]

\[= \frac{a - a + 1}{a} = \frac{1}{a}\]

\[= \frac{2a^{3} + 2a^{2} - 10a - 10}{(a + 1)^{2}(a + 2)} - 1 =\]

\[= \frac{2a^{2}(a + 1) - 10 \cdot (a + 1)}{(a + 1)^{2}(a + 2)} - 1 =\]

\[= \frac{2a^{2} - 10 - (a + 1)(a + 2)}{(a + 1)(a + 2)} =\]

\[= \frac{2a^{2} - 10 - a^{2} - 3a - 2}{(a + 1)(a + 2)} =\]

\[= \frac{a² - 3a - 12}{a² + 3a + 2}\]

\[= \frac{9x + 3}{\left( 9x^{2} + 3x + 1 \right)(3x + 1)} =\]

\[= \frac{3}{9x² + 3x + 1}.\]