Решебник по алгебре 9 класс Макарычев Задание 591

 

\[\boxed{\text{591}\text{\ (591)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x_{1} = 2;\ \ x_{2} = 9\]

\[d = x_{2} - x_{1} = 7\ \]

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

\[x_{n} = 2 + 7n - 7 = 7n - 5\]

\[\textbf{а)}\ 156?\]

\[7n - 5 = 156\]

\[7n = 161\]

\[n = 23 - целое\ число,\ тогда:\]

\[a_{23} = 156.\]

\[\textbf{б)}\ 7n - 5 = 295\]

\[7n = 295\ \ \]

\[n = 42\frac{6}{7} - не\ целое\ число,\]

\[значит,\ арифметическая\ \]

\[прогрессия\ не\ содержит\ \]

\[число\ 295.\]

\[\boxed{\text{591.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x_{n} = x_{1} \cdot q^{n - 1};\]

\[\textbf{а)}\ x_{1} = 16;\ \ q = \frac{1}{2}:\]

\[x_{7} = x_{1} \cdot q^{6} = 16 \cdot \left( \frac{1}{2} \right)^{6} =\]

\[= 2^{4} \cdot 2^{- 6} = 2^{- 2} = \frac{1}{4}.\]

\[\textbf{б)}\ x_{1} = - 810;\ \ q = \frac{1}{3}:\]

\[x_{8} = x_{1} \cdot q^{7} = - 810 \cdot \left( \frac{1}{3} \right)^{7} =\]

\[= - 10 \cdot 3^{4} \cdot 3^{- 7} = - \frac{10}{27}.\]

\[\textbf{в)}\ x_{1} = \sqrt{2};\ \ q = - \sqrt{2}:\]

\[x_{10} = x_{1} \cdot q^{9} = \sqrt{2} \cdot \left( - \sqrt{2} \right)^{9} =\]

\[= - \left( \sqrt{2} \right)^{10} = - 2^{5} = - 32.\]

\[\textbf{г)}\ x_{1} = - 125;\ \ q = 0,2:\]

\[x_{6} = x_{1} \cdot q^{5} = - 125 \cdot (0,2)^{5} =\]

\[= - 5^{3} \cdot \frac{1}{5^{5}} = - 5^{3} \cdot 5^{- 5} =\]

\[= - 5^{- 2} = - \frac{1}{25}.\]

\[\textbf{д)}\ x_{1} = \frac{3}{4};\ \ \ q = \frac{2}{3}:\]

\[x_{5} = x_{1} \cdot q^{4} = \frac{3}{4} \cdot \left( \frac{2}{3} \right)^{4} =\]

\[= \frac{3}{4} \cdot \frac{16}{81} = \frac{4}{27}.\]

\[\textbf{е)}\ x_{1} = 1,8 = \frac{18}{10} = \frac{9}{5};\ \ \ q = \frac{\sqrt{3}}{3}:\]

\[x_{4} = x_{1} \cdot q^{3} = \frac{9}{5} \cdot \left( \frac{\sqrt{3}}{3} \right)^{3} =\]

\[= \frac{9}{5} \cdot \frac{3\sqrt{3}}{27} = \frac{\sqrt{3}}{5}.\]