Решебник по алгебре 8 класс Мерзляк Задание 558

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\[1)\frac{\sqrt{5}}{\sqrt{5} - 2} = \frac{\sqrt{5} \cdot \left( \sqrt{5} + 2 \right)}{\left( \sqrt{5} + 2 \right)\left( \sqrt{5} - 2 \right)} =\]

\[= \frac{5 + 2\sqrt{5}}{5 - 4} = 5 + 2\sqrt{5}\]

\[2)\ \frac{8}{\sqrt{10} - \sqrt{2}} =\]

\[= \frac{8 \cdot \left( \sqrt{10} + \sqrt{2} \right)}{\left( \sqrt{10} - \sqrt{2} \right)\left( \sqrt{10} + \sqrt{2} \right)} =\]

\[= \sqrt{10} + \sqrt{2}\]

\[3)\ \frac{9}{\sqrt{x} + \sqrt{y}} =\]

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

\[= \frac{9 \cdot \left( \sqrt{x} - \sqrt{y} \right)}{x - y}\]

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

\[= \frac{\left( 2 - \sqrt{2} \right)^{2}}{2} = \frac{4 - 4\sqrt{2} + 6}{2} =\]

\[= \frac{6 - 4\sqrt{2}}{2} = 3 - 2\sqrt{2}\]