Решебник по алгебре 9 класс Макарычев Задание 681

 

\[\boxed{\text{681}\text{\ (681)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Дано:\left( y_{n} \right) - арифметическая\ прогрессия.\]

\[Доказать:\]

\[\textbf{а)}\ y_{2} + y_{7} = y_{4} + y_{5}\]

\[y_{2} = y_{1} + d,\]

\[y_{7} = y_{1} + 6d,\]

\[y_{4} = y_{1} + 3d\]

\[y_{5} = y_{1} + 4d\]

\[y_{1} + d + y_{1} + 6d = y_{1} + 3d + y_{1} + 4d\]

\[2y_{1} + 7d = 2y_{1} + 7d \Longrightarrow верно.\]

\[\textbf{б)}\ y_{n - 5} + y_{n + 10} = y_{n} + y_{n + 5},\ \ где\ \ n > 5.\]

\[y_{n + 10} = y_{1} + d(n + 9)\]

\[y_{n - 5} = y_{1} + d(n - 6)\]

\[y_{n} = y_{1} + d(n - 1)\]

\[y_{n + 5} = y_{1} + d(n + 4)\]

\[y_{1} + d(n - 6) + y_{1} + d(n + 9) = y_{1} + d(n - 1) + y_{1} + d(n + 4)\]

\[2y_{1} + d(2n + 3) = 2y_{1} + d(2n + 3) \Longrightarrow ч.т.д.\]

\[\boxed{\text{681.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{2 - 3x^{2}}{x^{3}} = \frac{2}{x^{3}} - \frac{3x^{2}}{x^{3}} = \frac{2}{x^{3}} - \frac{3}{x}\]

\[при\ x = - \frac{1}{2}:\]

\[2 \cdot ( - 2)^{3} - 3 \cdot ( - 2) =\]

\[= - 16 + 6 = - 10.\ \]

\[\textbf{б)}\ \frac{1 - m^{2}}{3m^{2} - m}\]

\[при\ m = \frac{2}{3}:\]

\[\frac{1 - \left( \frac{2}{3} \right)^{2}}{3 \cdot \left( \frac{2}{3} \right)^{2} - \frac{2}{3}} = \frac{5}{9} \cdot \frac{3}{2} = \frac{5}{6}.\]

\[\textbf{в)}\ \frac{10x^{2} - 5y^{2}}{x + y}\]

\[при\ \ x = 1,4;\ \ \ y = - 1,6:\]

\[\frac{10 \cdot {1,4}^{2} - 5 \cdot ( - 1,6)^{2}}{1,4 - 1,6} =\]

\[= - \frac{19,6 - 12,8}{0,2} = - 34.\]

\[\textbf{г)}\ \frac{\text{abc}}{a(b - c)} = \frac{\text{bc}}{b - c}\]

\[при\ \ a = 1,5;\ \ \ b = 10:\]

\[\frac{10 \cdot ( - 2)}{10 - ( - 2)} = - \frac{20}{12} = - \frac{5}{3} = - 1\frac{2}{3}.\]