ГДЗ по алгебре 8 класс Мерзляк Задание 256

\[\boxed{\text{256\ (256).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

\[2)\ 3a^{- 1} + ab^{- 2} = \frac{3^{\backslash b^{2}}}{a} + \frac{a^{\backslash a}}{b^{2}} =\]

\[= \frac{3b^{2} + a^{2}}{ab^{2}}\]

\[3)\ m^{2}n^{2} \cdot \left( m^{- 3} - n^{- 3} \right) =\]

\[= m^{2}n^{2}\left( \frac{1^{\backslash n^{3}}}{m^{3}} - \frac{1^{\backslash m^{3}}}{n^{3}} \right) =\]

\[= \frac{m^{2}n^{2}(n^{3} - m^{3})}{n^{3}m^{3}} = \frac{n^{3} - m^{3}}{\text{nm}}\]

\[4)\ (a + b)^{- 1} \cdot \left( a^{- 1} + b^{- 1} \right) =\]

\[= \frac{1}{a + b} \cdot \left( \frac{1^{\backslash b}}{a} + \frac{1^{\backslash a}}{b} \right) =\]

\[= \frac{(a + b)}{(a + b)\text{ab}} = \frac{1}{\text{ab}}\]

\[5)\ \left( c^{- 2} - d^{- 2} \right)\ :(c + d) =\]

\[= \left( \frac{1^{\backslash d^{2}}}{c^{2}} - \frac{1^{\backslash c^{2}}}{d^{2}} \right)\ :(c + d) =\]

\[= \frac{(d - c)(c + d)}{c^{2}d^{2}(c + d)} = \frac{d - c}{c^{2}d^{2}}\]

\[6)\ \left( xy^{- 2} + x^{- 2}y \right) \cdot\]

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

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

\[\cdot \frac{x}{x^{2} - xy + y^{2}} =\]

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

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

\[= \frac{x + y}{xy^{2}}\]