Решебник по алгебре 8 класс Мерзляк ФГОС Задание 167

\[\boxed{\text{167\ (167).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{25 - 5a + 5b - ab}{25 + 5a - 5b - ab} \cdot\]

\[\cdot \frac{ab - 5a - 5b + 25}{ab + 5a + 5b + 25} =\]

\[= \frac{\left( 5(5 - a) + b(5 - a) \right)\left( a(b - 5) - 5(b - 5) \right)}{\left( 5(5 + a) - b(5 + a) \right)\left( a(b + 5) + 5(b + 5) \right)} =\]

\[= \frac{(5 - a)(5 + b)(b - 5)(a - 5)}{(5 + a)(5 - b)(b + 5)(a + 5)} =\]

\[= \frac{(a - 5)^{2}}{(a + 5)^{2}}\]

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

\[:\frac{a^{2} - ab + 4a - 4b}{a^{2} - 16} =\]

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

\[= 1\]

\[\boxed{\text{167.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\frac{a^{2} + a}{2a - 12} \cdot \frac{6a + 6}{2a + 12}\ :\]

\[:\frac{9a^{3} + 18a^{2} + 9a}{a^{2} - 36} = \frac{1}{6}\]

\[Упростим\ левую\ часть\ \]

\[тождества:\]

\[\frac{a(a + 1) \cdot 6(a + 1) \cdot (a - 6)(a + 6)}{2(a - 6) \cdot 2(a + 6) \cdot 9a(a + 1)^{2}} =\]

\[= \frac{1}{6}\]

\[\frac{3}{18} = \frac{1}{6}\]

\[\frac{1}{6} = \frac{1}{6}.\]

\[Тождество\ доказано.\]