Решебник по алгебре 9 класс Мерзляк Задание 865

\[\boxed{\mathbf{865\ (865).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{7^{9}}{7^{10}} = \frac{1}{7}\]

\[2)\ \frac{125^{3}}{25^{4}} = \frac{\left( 5^{3} \right)^{3}}{\left( 5^{2} \right)^{4}} = \frac{5^{9}}{5^{8}} = 5^{1} = 5\]

\[3)\ \frac{32^{5}}{64^{4}} = \frac{\left( 2^{5} \right)^{5}}{\left( 2^{6} \right)^{4}} = \frac{2^{25}}{2^{24}} = 2^{1} = 2\]

\[4)\ \frac{39^{8}}{3^{10} \cdot 13^{7}} = \frac{3^{8} \cdot 13^{8}}{3^{10} \cdot 13^{7}} =\]

\[= \frac{13}{3^{2}} = \frac{13}{9}\]

\[\boxed{\mathbf{865.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[b_{1} = 6\sqrt{3},\ \ S = 9 \cdot \left( \sqrt{3} + 1 \right),\]

\[\text{\ \ }|q| < 1\]

\[S = \frac{b_{1}}{1 - q} \Longrightarrow \ 1 - q = \frac{b_{1}}{S} \Longrightarrow \text{\ \ }\]

\[q = 1 - \frac{b_{1}}{S} = 1 - \frac{6\sqrt{3}}{9 \cdot \left( \sqrt{3} + 1 \right)} =\]

\[= \frac{9 \cdot \left( \sqrt{3} + 1 \right) - 6\sqrt{3}}{9 \cdot \left( \sqrt{3} + 1 \right)} =\]

\[= \frac{9\sqrt{3} + 9 - 6\sqrt{3}}{9 \cdot \left( \sqrt{3} + 1 \right)} =\]

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

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

\[Ответ:q = \frac{\sqrt{3}}{3}.\]