Решебник по алгебре 9 класс Макарычев Задание 910

\[\boxed{\text{910\ (910).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{2}{x^{2} - 3x} - \frac{1}{x^{2} + 3x} - \frac{x + 1}{x^{2} - 9} =\]

\[= \frac{2x^{2} + 6x - x^{2} + 3x}{x^{4} - 9x^{2}} - \frac{x + 1}{x^{2} - 9} =\]

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

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

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

\[= \frac{x(9 - x^{2})}{x²(x^{2} - 9)} = - \frac{1}{x};\]

\[\textbf{б)}\ \frac{2y + 1}{y^{2} + 3y} + \frac{y + 2}{3y - y^{2}} - \frac{1}{y} =\]

\[= \frac{10y²}{y²(9 - y^{2})} = \frac{10}{9 - y²}\]

\[= \frac{a² + 2a + 4}{(a - 2)(a^{2} + 2a + 4)} = \frac{1}{a - 2}\]

\[= \frac{5 \cdot (2b + 3)}{(2b + 3)\left( 4b^{2} - 6b + 9 \right)} =\]

\[= \frac{5}{4b² - 6b + 9}.\]