Решебник по алгебре 8 класс Мерзляк Задание 559

\[\boxed{\mathbf{559\ (559).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{1}{5 - 2\sqrt{6}} + \frac{1}{5 + 2\sqrt{6}} = 10\]

\[\frac{5 + 2\sqrt{6} + 5 - 2\sqrt{6}}{\left( 5 - 2\sqrt{6} \right)\left( 5 + 2\sqrt{6} \right)} = 10\]

\[\frac{10}{25 - 24} = 10\]

\[10 = 10.\]

\[2)\ \frac{2}{3\sqrt{2} + 4} - \frac{2}{3\sqrt{2} - 4} = - 8\]

\[\frac{6\sqrt{2} - 8 - 6\sqrt{2} - 8}{\left( 3\sqrt{2} + 4 \right)\left( 3\sqrt{2} - 4 \right)} = - 8\]

\[\frac{- 16}{18 - 16} = - 8\]

\[- \frac{16}{2} = - 8\]

\[- 8 = - 8.\]

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

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

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

\[\frac{4\sqrt{2}}{1} = 4\sqrt{2}\]

\[4\sqrt{2} = 4\sqrt{2}.\]