Решебник по алгебре 8 класс Мерзляк Задание 182

\[\boxed{\text{182\ (182).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[Преобразуем\ левую\ часть\ равенства:\]

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

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

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

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

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

\[Тождество\ доказано.\]

\[2)\ \left( \frac{8a}{4 - a^{2}} - \frac{a - 2}{a + 2} \right)\ :\frac{a + 2}{a} +\]

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

\[Упростим\ левую\ часть\ равенства:\]

\[\left( \frac{8}{(2 - a)(2 + a)} - \frac{a - 2^{\backslash 2 - a}}{a + 2} \right)\ :\]

\[:\frac{a + 2}{a} + \frac{2}{a - 2} = - 1\]

\[\frac{8a + a^{2} - 4a + 4}{4 - a^{2}}\ :\frac{a + 2}{a} +\]

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

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

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

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

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

\[- 1 = - 1.\]

\[Тождество\ доказано.\]

\[3)\ \left( \frac{3}{36 - c^{2}} + \frac{1}{c^{2} - 12c + 36} \right) \cdot\]

\[\cdot \frac{(c - 6)^{2}}{2} + \frac{3c}{c + 6} = 2\]

\[\left( \frac{3^{\backslash 6 - c}}{(6 - c)(6 + c)} + \frac{1^{\backslash 6 + c}}{(6 - c)^{2}} \right) \cdot\]

\[\cdot \frac{(c - 6)^{2}}{2} + \frac{3c}{c + 6} = 2\]

\[Упростим\ левую\ часть\ \]

\[равенства:\]

\[\frac{18 - 3c + c + 6}{(c - 6)^{2}(c + 6)} \cdot \frac{(c - 6)^{2}}{2} +\]

\[+ \frac{3c}{c + 6} = 2\]

\[\frac{2 \cdot (12 - c)(c - 6)^{2}}{(c - 6)^{2}(c + 6) \cdot 2} + \frac{3c}{c + 6} = 2\]

\[\frac{12 - c}{c + 6} + \frac{3c}{c + 6} = 2\]

\[\frac{12 - c + 3c}{c + 6} = 2\]

\[\frac{2 \cdot (6 + c)}{(c + 6)} = 2\]

\[2 = 2.\]

\[Тождество\ доказано.\]