Решебник по алгебре 8 класс Мерзляк ФГОС Задание 557

 

\[\boxed{\mathbf{557\ (557).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

\[2)\ \frac{4}{\sqrt{7} + \sqrt{3}} =\]

\[= \frac{4 \cdot \left( \sqrt{7} - \sqrt{3} \right)}{\left( \sqrt{7} + \sqrt{3} \right)\left( \sqrt{7} - \sqrt{3} \right)} =\]

\[= \frac{4 \cdot \left( \sqrt{7} - \sqrt{3} \right)}{7 - 3} = \sqrt{7} - \sqrt{3}\]

\[3)\ \frac{15}{\sqrt{15} - \sqrt{12}} =\]

\[= \frac{15 \cdot \left( \sqrt{15} + \sqrt{12} \right)}{\left( \sqrt{15} - \sqrt{12} \right)\left( \sqrt{15} + \sqrt{12} \right)} =\]

\[= \frac{15 \cdot \left( \sqrt{15} + \sqrt{12} \right)}{15 - 12} =\]

\[= 5 \cdot (\sqrt{15} + \sqrt{12})\]

\[4)\ \frac{19}{2\sqrt{5} - 1} =\]

\[= \frac{19 \cdot \left( 2\sqrt{5} + 1 \right)}{\left( 2\sqrt{5} - 1 \right)\left( 2\sqrt{5} + 1 \right)} =\]

\[= \frac{19 \cdot \left( 2\sqrt{5} + 1 \right)}{20 - 1} = 2\sqrt{5} + 1\]

\[5)\ \frac{1}{\sqrt{a} - \sqrt{b}} =\]

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

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

\[6)\ \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{\left( \sqrt{3} + 1 \right)^{2}}{\left( \sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right)} =\]

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

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

\[\boxed{\mathbf{5}\mathbf{5}\mathbf{7}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{9a} + \sqrt{25a} - \sqrt{49a} =\]

\[= 3\sqrt{a} + 5\sqrt{a} - 7\sqrt{a} = \sqrt{a};\]

\[2)\ \sqrt{64b} - \frac{1}{6}\sqrt{36b} = 8\sqrt{b} - \sqrt{b} =\]

\[= 7\sqrt{b}\]

\[= 0,4\sqrt{c} - 1,2\sqrt{c} + 0,3\sqrt{c} =\]

\[= - 0,5\sqrt{c};\]

\[= 4\sqrt{m} + 10\sqrt{m} - 1,8\sqrt{m} =\]

\[= 12,2\sqrt{m}\]