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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 744
Задание 744
\[\boxed{\mathbf{744\ (744).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ a_{n} = 6 + 7n\]
\[a_{n + 1} - a_{n} = 6 + 7 \cdot (n + 1) -\]
\[- (6 + 7n) = 6 + 7n + 7 - 6 -\]
\[- 7n = 7 \Longrightarrow да,\]
\[d = 7;\ \ \ \ a_{1} = 6 + 7 \cdot 1 = 13.\]
\[2)\ a_{n} = \frac{2n - 1}{5}\]
\[a_{n + 1} - a_{n} = \frac{2 \cdot (n + 1) - 1}{5} -\]
\[- \frac{2n - 1}{5} =\]
\[= \frac{2n + 2 - 1 - 2n + 1}{5} =\]
\[= \frac{2}{5} \Longrightarrow да,\ \]
\[d = \frac{2}{5};\ \ \ a_{1} = (2 \cdot 1 - 1)\ :5 = \frac{1}{5}.\]
\[3)\ a_{n} = \frac{1}{n} + 2\]
\[a_{n + 1} - a_{n} = \frac{1}{n + 1} + 2 - \frac{1}{n} -\]
\[- 2 = \frac{1}{n + 1} - \frac{1}{n} = \frac{n - n - 1}{n(n + 1)} =\]
\[= - \frac{1}{n(n + 1)}\]
\[Не\ является.\]
\[Ответ:1)\ да:\ \ d = 7;\ a_{1} = 13;\ \]
\[\ 2)\ да:\ \ d = \frac{2}{5};\ \ a_{1} = \frac{1}{5};\ \ 3)\ нет.\]
\[\boxed{\mathbf{744.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[\left\{ \begin{matrix} b_{2}b_{4} = 36\ \ \ \\ b_{3} + b_{5} = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} b_{1}q \cdot b_{1}q^{3} = 36 \\ b_{1}q^{2} + b_{1}q^{4} = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} b_{1}^{2} \cdot q^{4} = 36\ \ \ \ \ \ \ \ \ \\ b_{1}q^{2}\left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} b_{1}q^{2} = \pm 6\ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{2}\left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} b_{1}q^{2} = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6 \cdot \left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} b_{1}q^{2} = 6\ \ \ \ \\ 1 + q^{2} = \frac{8}{6} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} b_{1}q^{2} = 6 \\ q^{2} = \frac{1}{3}\text{\ \ \ \ } \\ \end{matrix} \right.\ \ \]
\[\text{\ \ \ }\left\{ \begin{matrix} b_{1}q^{2} = 6 \\ q = \pm \sqrt{\frac{1}{3}\ } \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\ \left\{ \begin{matrix} b_{1} = \frac{6}{q^{2}}\text{\ \ \ } \\ q = \pm \sqrt{\frac{1}{3}\ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} b_{1} = \frac{6}{\left( \sqrt{\frac{1}{3}} \right)^{2}} \\ q = \pm \sqrt{\frac{1}{3}\ }\text{\ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} b_{1} = 18\ \ \ \\ q = \pm \sqrt{\frac{1}{3}\ } \\ \end{matrix} \right.\ \]
\[или\ \ \ \ \ \left\{ \begin{matrix} b_{1}q^{2} = - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 6 \cdot \left( 1 + q^{2} \right) = 8 \\ \end{matrix} \right.\ \Longrightarrow \ \]
\[\Longrightarrow \ 1 + q^{2} = - \frac{8}{6}\text{\ \ }\]
\[q^{2} = - 2\frac{1}{3} < 0 \Longrightarrow не\ \]
\[удовлетворяет.\]
\[При\ b_{1} = 18 \Longrightarrow \ \ q = \sqrt{\frac{1}{3}\ }\ \]
\[S = \frac{18}{1 - \sqrt{\frac{1}{3}\ }} = \frac{18}{\frac{1 - \sqrt{3}}{\sqrt{3}}} =\]
\[= \frac{18\sqrt{3}}{1 - \sqrt{3}} = \frac{18\sqrt{3} \cdot \left( 1 + \sqrt{3} \right)}{\left( 1 + \sqrt{3} \right)\left( 1 - \sqrt{3} \right)} =\]
\[= \frac{18\sqrt{3} \cdot \left( 1 + \sqrt{3} \right)}{- 1 + 3} =\]
\[= 9\sqrt{3} \cdot \left( 1 - \sqrt{3} \right) = 9\sqrt{3} + 27.\]
\[При\ b_{1} = 18 \Longrightarrow \ \ q = - \sqrt{\frac{1}{3}\ }\text{\ \ }\]
\[S = \frac{18}{1 + \sqrt{\frac{1}{3}\ }} = \frac{18}{\frac{1 + \sqrt{3}}{\sqrt{3}}} =\]
\[= \frac{18\sqrt{3}}{1 + \sqrt{3}} = \frac{18\sqrt{3} \cdot \left( 1 - \sqrt{3} \right)}{\left( 1 + \sqrt{3} \right)\left( 1 - \sqrt{3} \right)} =\]
\[= \frac{18\sqrt{3} \cdot \left( 1 - \sqrt{3} \right)}{1 - 3} =\]
\[= - 9\sqrt{3} \cdot \left( 1 - \sqrt{3} \right) =\]
\[= - 9\sqrt{3} + 27.\]
\[Ответ:\ S = 27 \pm 9\sqrt{3}.\]
