Решебник по алгебре 9 класс Мерзляк Задание 744

\[\boxed{\mathbf{744\ (744).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ a_{n} = 6 + 7n\]

\[a_{n + 1} - a_{n} = 6 + 7 \cdot (n + 1) -\]

\[- (6 + 7n) = 6 + 7n + 7 - 6 -\]

\[- 7n = 7 \Longrightarrow да,\]

\[d = 7;\ \ \ \ a_{1} = 6 + 7 \cdot 1 = 13.\]

\[2)\ a_{n} = \frac{2n - 1}{5}\]

\[a_{n + 1} - a_{n} = \frac{2 \cdot (n + 1) - 1}{5} -\]

\[- \frac{2n - 1}{5} =\]

\[= \frac{2n + 2 - 1 - 2n + 1}{5} =\]

\[= \frac{2}{5} \Longrightarrow да,\ \]

\[d = \frac{2}{5};\ \ \ a_{1} = (2 \cdot 1 - 1)\ :5 = \frac{1}{5}.\]

\[3)\ a_{n} = \frac{1}{n} + 2\]

\[a_{n + 1} - a_{n} = \frac{1}{n + 1} + 2 - \frac{1}{n} -\]

\[- 2 = \frac{1}{n + 1} - \frac{1}{n} = \frac{n - n - 1}{n(n + 1)} =\]

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

\[Не\ является.\]

\[Ответ:1)\ да:\ \ d = 7;\ a_{1} = 13;\ \]

\[\ 2)\ да:\ \ d = \frac{2}{5};\ \ a_{1} = \frac{1}{5};\ \ 3)\ нет.\]

\[\boxed{\mathbf{744.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\left\{ \begin{matrix} b_{2}b_{4} = 36\ \ \ \\ b_{3} + b_{5} = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} b_{1}q \cdot b_{1}q^{3} = 36 \\ b_{1}q^{2} + b_{1}q^{4} = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b_{1}^{2} \cdot q^{4} = 36\ \ \ \ \ \ \ \ \ \\ b_{1}q^{2}\left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b_{1}q^{2} = \pm 6\ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{2}\left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b_{1}q^{2} = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6 \cdot \left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} b_{1}q^{2} = 6\ \ \ \ \\ 1 + q^{2} = \frac{8}{6} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} b_{1}q^{2} = 6 \\ q^{2} = \frac{1}{3}\text{\ \ \ \ } \\ \end{matrix} \right.\ \ \]

\[\text{\ \ \ }\left\{ \begin{matrix} b_{1}q^{2} = 6 \\ q = \pm \sqrt{\frac{1}{3}\ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\ \left\{ \begin{matrix} b_{1} = \frac{6}{q^{2}}\text{\ \ \ } \\ q = \pm \sqrt{\frac{1}{3}\ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} b_{1} = \frac{6}{\left( \sqrt{\frac{1}{3}} \right)^{2}} \\ q = \pm \sqrt{\frac{1}{3}\ }\text{\ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} b_{1} = 18\ \ \ \\ q = \pm \sqrt{\frac{1}{3}\ } \\ \end{matrix} \right.\ \]

\[или\ \ \ \ \ \left\{ \begin{matrix} b_{1}q^{2} = - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 6 \cdot \left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \Longrightarrow \ \]

\[\Longrightarrow \ 1 + q^{2} = - \frac{8}{6}\text{\ \ }\]

\[q^{2} = - 2\frac{1}{3} < 0 \Longrightarrow не\ \]

\[удовлетворяет.\]

\[При\ b_{1} = 18 \Longrightarrow \ \ q = \sqrt{\frac{1}{3}\ }\ \]

\[S = \frac{18}{1 - \sqrt{\frac{1}{3}\ }} = \frac{18}{\frac{1 - \sqrt{3}}{\sqrt{3}}} =\]

\[= \frac{18\sqrt{3}}{1 - \sqrt{3}} = \frac{18\sqrt{3} \cdot \left( 1 + \sqrt{3} \right)}{\left( 1 + \sqrt{3} \right)\left( 1 - \sqrt{3} \right)} =\]

\[= \frac{18\sqrt{3} \cdot \left( 1 + \sqrt{3} \right)}{- 1 + 3} =\]

\[= 9\sqrt{3} \cdot \left( 1 - \sqrt{3} \right) = 9\sqrt{3} + 27.\]

\[При\ b_{1} = 18 \Longrightarrow \ \ q = - \sqrt{\frac{1}{3}\ }\text{\ \ }\]

\[S = \frac{18}{1 + \sqrt{\frac{1}{3}\ }} = \frac{18}{\frac{1 + \sqrt{3}}{\sqrt{3}}} =\]

\[= \frac{18\sqrt{3}}{1 + \sqrt{3}} = \frac{18\sqrt{3} \cdot \left( 1 - \sqrt{3} \right)}{\left( 1 + \sqrt{3} \right)\left( 1 - \sqrt{3} \right)} =\]

\[= \frac{18\sqrt{3} \cdot \left( 1 - \sqrt{3} \right)}{1 - 3} =\]

\[= - 9\sqrt{3} \cdot \left( 1 - \sqrt{3} \right) =\]

\[= - 9\sqrt{3} + 27.\]

\[Ответ:\ S = 27 \pm 9\sqrt{3}.\]