ГДЗ по алгебре 8 класс Макарычев Задание 149

\[\boxed{\text{149\ (149).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( \frac{x}{x + 1} + 1^{\backslash x + 1} \right) \cdot \frac{1 + x}{2x - 1} =\]

\[= \frac{x}{x + 1} \cdot \frac{1 + x}{2x - 1} + \frac{1 + x}{2x - 1} =\]

\[= \frac{x}{2x - 1} + \frac{1 + x}{2x - 1} =\]

\[= \frac{x + 1 + x}{2x - 1} = \frac{2x + 1}{2x - 1}\]

\[\textbf{б)}\ \frac{5y^{2}}{1 - y^{2}}\ \ :\left( 1^{\backslash 1 - y} - \frac{1}{1 - y} \right) =\]

\[= \frac{5y^{2}}{1 - y^{2}}\ :\frac{1 - y - 1}{1 - y} =\]

\[= \frac{5y^{2}}{(1 - y)(1 + y)} \cdot \frac{1 - y}{- y} =\]

\[= - \frac{5y}{1 + y}\]

\[\textbf{в)}\ \left( \frac{4a}{2 - a} - a^{\backslash 2 - a} \right)\ :\ \frac{a + 2}{a - 2} =\]

\[= \frac{4a - 2a + a^{2}}{2 - a} \cdot \frac{a - 2}{a + 2} =\]

\[= - \frac{2a + a^{2}}{a - 2} \cdot \frac{a - 2}{a + 2} =\]

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

\[\textbf{г)}\ \frac{x - 2}{x - 3} \cdot \left( x^{\backslash x - 2} + \frac{x}{2 - x} \right) =\]

\[= \frac{x(x - 2)}{x - 3} - \frac{x}{x - 2} \cdot \frac{x - 2}{x - 3} =\]

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

\[= \frac{x(x - 2 - 1)}{x - 3} = \frac{x(x - 3)}{x - 3} = x\]