ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 529

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

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

\[= 3 - 2\sqrt{2};\]

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

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

\[S = \frac{b_{1}}{1 - q} = \frac{3 + 2\sqrt{2}}{1 - 3 + 2\sqrt{2}} =\]

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

\[= \frac{5\sqrt{2} + 7}{2}.\]

\[2)\ \frac{\sqrt{3} + 1}{\sqrt{3} - 1};\ \ 1;\ \ \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\]

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

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

\[= \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} =\]

\[= 2 + \sqrt{3};\]

\[S = \frac{b_{1}}{1 - q} = \frac{2 + \sqrt{3}}{1 - \sqrt{3} + 2} =\]

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

\[= \frac{9 + 5\sqrt{3}}{9 - 3} = \frac{9 + 5\sqrt{3}}{6}.\]