Решебник по алгебре 9 класс Макарычев Задание 596

 

\[\boxed{\text{596}\text{\ (596)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[a^{2} = x;\ \ b^{2} = x + d;\ \ \]

\[c² = x + 2d;\]

\[\frac{1}{a + c} =\]

\[= \frac{1}{b + c} + \left( \frac{1}{a + c} - \frac{1}{b + c} \right) =\]

\[= \frac{1}{b + c} + \frac{b - a}{(a + c)(b + c)} =\]

\[= \frac{1}{b + c} + \frac{b^{2} - a^{2}}{(a + c)(b + c)(a + b)} =\]

\[= \left( \frac{1}{b + c} \right) + \frac{d}{(a + c)(b + c)(a + b)};\]

\[\frac{1}{a + b} =\]

\[= \frac{1}{b + c} + \left( \frac{1}{a + b} - \frac{1}{b + c} \right) =\]

\[= \frac{1}{b + c} + \frac{b + c - a - b}{(a + b)(b + c)} =\]

\[= \frac{1}{b + c} + \frac{c - a}{(a + b)(b + c)} =\]

\[= \frac{1}{b + c} + \frac{c^{2} - a^{2}}{(a + c)(b + c)(a + b)} =\]

\[= \frac{1}{b + c} + \frac{2d}{(a + c)(b + c)(a + b)}.\]

\[Что\ и\ требовалось\ доказать.\]

\[\boxed{\text{596.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ b_{6} = 3,\ \ q = 3,\ \ \]

\[b_{6} = b_{1} \cdot q^{5},\ \ 3 = b_{1} \cdot 3^{5},\]

\[b_{1} = \frac{3}{3^{5}} = 3^{- 4} = \frac{1}{81};\ \]

\[\textbf{б)}\ b_{5} = 17\frac{1}{2} = \frac{35}{2};\ \ \ \]

\[\text{\ \ }q = - 2\frac{1}{2} = - \frac{5}{2};\ \ \ \ \ \ \]

\[\text{\ \ \ }b_{5} = b_{1} \cdot q^{4},\ \]

\[\frac{35}{2} = b_{1} \cdot \left( - \frac{5}{2} \right)^{4},\ \ \]

\[b_{1} = \frac{35}{2} \cdot \frac{2^{4}}{5^{4}} = \frac{2³ \cdot 7}{5³} = \frac{56}{125}.\]

\[Ответ:а)\ \frac{1}{81};\ \ б)\frac{56}{125}.\]