ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 92

\[\boxed{\text{92.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 1^{\backslash a - b} - \frac{a + b}{a - b} =\]

\[= \frac{a - b - a - b}{a - b} = \frac{- 2b}{a - b} =\]

\[= \frac{2b}{b - a}\]

\[\textbf{б)}\ \frac{a^{2} + b^{2}}{a - b} - a^{\backslash a - b} =\]

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

\[\textbf{в)}\ (m - n)^{\backslash m + n} + \frac{n^{2}}{m + n} =\]

\[= \frac{m^{2} - n^{2} + n^{2}}{m + n} = \frac{m^{2}}{m + n}\]

\[\textbf{г)}\ (a + b)^{\backslash a + b} - \frac{a^{2} + b^{2}}{a + b} =\]

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

\[= \frac{2ab}{a + b}\]

\[\textbf{д)}\ x^{\backslash x - 3} - \frac{9}{x - 3} - 3^{\backslash x - 3} =\]

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

\[= \frac{x^{2} - 6x}{x - 3}\]

\[\textbf{е)}\ {a^{2}}^{\backslash a^{2} - 1} - \frac{a^{4} + 1}{a^{2} - 1} + 1^{\backslash a^{2} - 1} =\]

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

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