ГДЗ по алгебре 9 класс Мерзляк Задание 902

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\[1)\ \sqrt{2};\ - 1;\frac{1}{\sqrt{2}};\ldots\ \]

\[q = - \frac{1}{\sqrt{2}}\text{\ \ }\]

\[S = \frac{\sqrt{2}}{1 + \frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{\frac{\sqrt{2} + 1}{\sqrt{2}}} =\]

\[= \frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{2} + 1} = \frac{2}{\sqrt{2} + 1} =\]

\[= \frac{2 \cdot \left( \sqrt{2} - 1 \right)}{\left( \sqrt{2} + 1 \right)\left( \sqrt{2} - 1 \right)} =\]

\[= \frac{2 \cdot \left( \sqrt{2} - 1 \right)}{2 - 1} =\]

\[= 2 \cdot \left( \sqrt{2} - 1 \right).\]

\[2)\ 3\sqrt{3};3;\sqrt{3};\ldots\ \]

\[\ q = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}};\ \ \ \ \ \]

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

\[= \frac{3\sqrt{3} \cdot \sqrt{3}}{\sqrt{3} - 1} = \frac{9}{\sqrt{3} - 1} =\]

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

\[= \frac{9 \cdot (\sqrt{3} + 1)}{3 - 1} = 4,5 \cdot \left( \sqrt{3} + 1 \right).\]

\[3)\ \frac{\sqrt{3} + 1}{\sqrt{3} - 1},\ 1,\ \frac{\sqrt{3} - 1}{\sqrt{3} + 1},\ldots\ \]

\[q = \frac{1 \cdot (\sqrt{3} - 1)}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\]

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

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

\[= \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\ :\frac{2}{\sqrt{3} + 1} =\]

\[= \frac{\left( \sqrt{3} + 1 \right)\left( \sqrt{3} + 1 \right)}{\left( \sqrt{3} - 1 \right) \cdot 2} =\]

\[= \frac{\left( 3 - 2\sqrt{3} + 1 \right)\left( \sqrt{3} + 1 \right)}{\left( \sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right) \cdot 2} =\]

\[= \frac{\left( 4 - 2\sqrt{3} \right)\left( \sqrt{3} + 1 \right)}{2 \cdot \left( \sqrt{3} - 1 \right)} =\]

\[= \frac{4\sqrt{3} + 4 + 2 \cdot 3 + 2\sqrt{3}}{4} =\]

\[= \frac{6\sqrt{3} + 10}{4} = \frac{3\sqrt{3} + 5}{2}\]