ГДЗ по алгебре 8 класс Макарычев Задание 252

\[\boxed{\text{252\ (252).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\frac{1}{z - a} + \frac{1}{z - b} = \frac{1}{a} + \frac{1}{b}\]

\[Доказать:\ \]

\[z = \frac{2}{\frac{1}{a} + \frac{1}{b}}.\]

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

\[= \frac{1}{a} + \frac{1}{b}\]

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

\[= \frac{1}{a} + \frac{1}{b}\]

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

\[= \frac{1}{a} + \frac{1}{b}\]

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

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

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

\[\frac{1}{a} + \frac{1}{b} = \frac{1}{a} + \frac{1}{b}\]

\[Что\ и\ требовалось\ доказать.\]