ГДЗ по алгебре 9 класс Макарычев Задание 729

\[\boxed{\text{729\ (729).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{2ab}{a^{2} - b^{2}} + \frac{a - b}{2a + 2b} =\]

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

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

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

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

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

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

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

\[= \ \frac{a}{a - b} - \frac{b}{a - b} = \frac{a - b}{a - b} = 1.\]

\[1)\ \frac{x}{(x - y)^{2}} - \frac{y}{x^{2} - y^{2}} =\]

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

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

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

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

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

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

\[3)\frac{y}{x - y} - \frac{x}{x - y} = \frac{y - x}{x - y} = - 1.\ \]