Страница 243 номер 903, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите сумму бесконечной геометрической прогрессии: 1) \(\sqrt{\frac{3}{2}},\ 1,\ \sqrt{\frac{2}{3}},\ \ldots\); 2) \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1},\ \frac{1}{2 - \sqrt{2}},\ \frac{1}{2},\ \ldots\) .
1) \(\sqrt{\frac{3}{2}},1,\ \sqrt{\frac{2}{3}},\ldots\)
\(b_{1} = \sqrt{\dfrac{3}{2}};\ q = \sqrt{\dfrac{2}{3}}\)
\(S = \dfrac{b_{1}}{1 - q} = \dfrac{\sqrt{\frac{3}{2}}}{1 - \sqrt{\dfrac{2}{3}}} = \dfrac{\sqrt{3}}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{\sqrt{3} - \sqrt{2}} = \dfrac{3}{\sqrt{2}} \cdot \dfrac{\sqrt{3} + \sqrt{2}}{3 - 2} =\)
\(= \dfrac{3 \cdot \sqrt{2}(\sqrt{3} + \sqrt{2})}{2} = \dfrac{3(\sqrt{6} + 2)}{2}\)
2) \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1},\frac{1}{2 - \sqrt{2}},\frac{1}{2},\ldots\)
\(b_{1} = \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}\)
\(q = \dfrac{1}{2 - \sqrt{2}}:\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1} = \dfrac{1}{\sqrt{2}\left( \sqrt{2} - 1 \right)} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} + 1} = \dfrac{1}{\sqrt{2}\left( \sqrt{2} + 1 \right)} =\)
\(= \dfrac{1}{2 + \sqrt{2}}\)
\(S = \dfrac{b_{1}}{1 - q} = \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}:\left( 1 - \dfrac{1}{2 + \sqrt{2}} \right) = \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1} \cdot \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} - 1} =\)
\(= \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1} \cdot \dfrac{\sqrt{2}\left( \sqrt{2} + 1 \right)}{\sqrt{2} + 1} = \dfrac{\sqrt{2}\left( \sqrt{2} + 1 \right)^{2}}{2 - 1} = \sqrt{2}\left( 2 + 2\sqrt{2} + 1 \right) =\)
\(= \sqrt{2}\left( 3 + 2\sqrt{2} \right) = 3\sqrt{2} + 4\)
Ответ: 1) \(S = \frac{3(\sqrt{6} + 2)}{2}\); 2) \(S = 3\sqrt{2} + 4\).
