Решебник по алгебре 8 класс Мерзляк ФГОС Задание 598

\[\boxed{\mathbf{598\ (598).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{\left( 1 - \sqrt{2} \right)^{2}} = \left| 1 - \sqrt{2} \right| =\]

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

\[2)\ \sqrt{\left( \sqrt{6} - \sqrt{7} \right)^{2}} = \left| \sqrt{6} - \sqrt{7} \right| =\]

\[= \sqrt{7} - \sqrt{6}\]

\[3)\ \sqrt{\left( 2\sqrt{5} - 3 \right)^{2}} = 2\sqrt{5} - 3\]

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

\[= \left| \sqrt{3} - 2 \right| + \left| 3 - \sqrt{3} \right| =\]

\[= 2 - \sqrt{3} + 3 - \sqrt{3} = 5 - 2\sqrt{3}\]

\[\boxed{\mathbf{5}\mathbf{9}\mathbf{8}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{8 + 2\sqrt{7}} = \sqrt{7 + 2\sqrt{7} + 1} =\]

\[= \sqrt{\left( \sqrt{7} + 1 \right)^{2}} = \sqrt{7} + 1\]

\[2)\ \sqrt{15 + 6\sqrt{6}} = \sqrt{9 + 6\sqrt{6} + 6} =\]

\[= \sqrt{\left( 3 + \sqrt{6} \right)^{2}} = 3 + \sqrt{6}\]

\[3)\ \sqrt{7 + 2\sqrt{10}} =\]

\[= \sqrt{5 + 2\sqrt{10} + 2} =\]

\[= \sqrt{(\sqrt{5} + \sqrt{2})²} = \sqrt{5} + \sqrt{2}\]