Решебник по алгебре 8 класс Мерзляк ФГОС Задание 559

 

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

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

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

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

\[10 = 10.\]

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

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

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

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

\[- 8 = - 8.\]

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

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

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

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

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

\[\boxed{\mathbf{5}\mathbf{5}\mathbf{9}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 8\sqrt{2} - \sqrt{32} = 8\sqrt{2} - 4\sqrt{2} =\]

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

\[2)\ 6\sqrt{3} - \sqrt{27} = 6\sqrt{3} - 3\sqrt{3} =\]

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

\[3)\ \sqrt{96} - 3\sqrt{6} = 4\sqrt{6} - 3\sqrt{6} =\]

\[= \sqrt{6};\]

\[4)\ 2\sqrt{500} - 8\sqrt{5} =\]

\[= 20\sqrt{5} - 8\sqrt{5} = 12\sqrt{5};\]

\[5)\ 5\sqrt{7} - \sqrt{700} - 0,5\sqrt{28} =\]

\[= 5\sqrt{7} - 10\sqrt{7} - \sqrt{7} = - 6\sqrt{7};\]

\[6)\ 2\sqrt{20} - \frac{1}{3}\sqrt{45} - 0,6\sqrt{125} =\]

\[= 4\sqrt{5} - \frac{1}{3} \cdot 3\sqrt{5} - 0,6 \cdot 5\sqrt{5} =\]

\[= 4\sqrt{5} - \sqrt{5} - 3\sqrt{5} = 0\]