Решебник по алгебре 8 класс Мерзляк Задание 769

\[\boxed{\mathbf{769\ (769).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{3 + \sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3} \cdot \left( \sqrt{3} + 1 \right)}{2\sqrt{3}} =\]

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

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

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

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

\[= \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}\ \]

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

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

\[= 2\sqrt{a} - \sqrt{2}\]

\[5)\ \frac{9a - b^{2}}{9a + 6b\sqrt{a} + b^{2}} =\]

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

\[= \frac{3\sqrt{a} - b}{3\sqrt{a} + b}\]

\[6)\ \frac{a\sqrt{a} - 8}{a + 2\sqrt{a} + 4} =\]

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

\[= \sqrt{a} - 2\]