ГДЗ по алгебре 9 класс Макарычев Задание 729

 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\boxed{\text{729.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x^{2} - 10x + 31\]

\[\textbf{а)}\ x² - 10x + 31 = - 5\]

\[x^{2} - 10x + 36 = 0\]

\[D = 100 - 144 = - 44 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:таких\ x\ не\ существует.\]

\[\textbf{б)}\ x² - 10x + 31 = 6\]

\[x^{2} - 10x + 25 = 0\]

\[(x - 5)^{2} = 0\]

\[x = 5\]

\[Ответ:при\ x = 5.\]

\[\textbf{в)}\ x² - 10x + 31 = 55\]

\[x^{2} - 10x - 24 = 0\]

\[D = 100 + 96 = 196\]

\[x_{1} = \frac{10 + 14}{2} = 12,\ \ \]

\[x_{2} = \frac{10 - 14}{2} = - 2.\]

\[Ответ:при\ x = 12;\ x = - 2.\]