Решебник по алгебре 8 класс Мерзляк Задание 161

\[\boxed{\text{161\ (161).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{1}{a^{2} - ab}\ :\frac{b}{b^{2} - a^{2}} =\]

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

\[если\ a = 2\frac{1}{3} = \frac{7}{3};\ \ b = - \frac{3}{7}:\]

\[\ \left( \frac{3}{7} - \frac{7}{3} \right)\ :\left( - \frac{7}{3} \cdot \frac{3}{7} \right) =\]

\[= \frac{9 - 49}{21}\ :( - 1) = \frac{40}{21} = 1\frac{19}{21}.\]

\[2)\ \frac{a^{2} + 4ab + 4b^{2}}{a^{2} - 9b^{2}}\ :\frac{3a + 6b}{2a - 6b} =\]

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

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

\[если\ a = 4;\ \ b = - 5:\ \ \]

\[\frac{2 \cdot (4 - 2 \cdot 5)}{3 \cdot (4 - 3 \cdot 5)} = \frac{2 \cdot ( - 6)}{3 \cdot ( - 11)} =\]

\[= \frac{12}{33} = \frac{4}{11}.\]