Решебник по алгебре 8 класс Мерзляк Задание 183

\[\boxed{\text{183\ (183).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left( \frac{b}{a^{2} - ab} - \frac{2}{a - b} - \frac{a}{b^{2} - ab} \right)\ :\]

\[:\frac{a^{2} - b^{2}}{4ab} = \frac{4}{a + b}\]

\[Упростим\ левую\ часть\ \]

\[равенства:\]

\[\left( \frac{b^{\backslash b}}{a(a - b)} - \frac{2^{\backslash ab}}{a - b} + \frac{a^{\backslash a}}{b(a - b)} \right)\ :\]

\[:\frac{a^{2} - b^{2}}{4ab} = \frac{4}{a + b}\]

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

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

\[\frac{4}{a + b} = \frac{4}{a + b}.\]

\[Тождество\ доказано.\]

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

\[\cdot \left( \frac{a^{\backslash a + b}}{(a - b)^{2}} + \frac{a^{\backslash b - a}}{b^{2} - a^{2}} \right) +\]

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

\[Упростим\ левую\ часть\ \]

\[равенства:\]

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

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

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

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

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

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

\[3 = 3.\]

\[Тождество\ доказано.\]