Решебник по алгебре 8 класс Мерзляк ФГОС Задание 161

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

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

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

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

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

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

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

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

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

\[если\ a = 4;\ \ b = - 5:\ \ \]

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

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

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

\[3x + \frac{1}{x} = - 4;\ \ \ \ \ \ \ \ \ \]

\[9x^{2} + \frac{1}{x^{2}} = ?\]

\[Возведем\ обе\ части\ равенства\ \]

\[в\ квадрат:\]

\[\left( 3x + \frac{1}{x} \right)^{2} = ( - 4)^{2}\]

\[9x^{2} + 6 + \frac{1}{x^{2}} = 16\]

\[9x^{2} + \frac{1}{x^{2}} = 16 - 6\]

\[9x^{2} + \frac{1}{x^{2}} = 10\]

\[Ответ:10.\]