Решебник по алгебре 8 класс Мерзляк ФГОС Задание 581

 

\[\boxed{\mathbf{581\ (581).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[y = \sqrt{x}\]

\[\boxed{\mathbf{5}\mathbf{8}\mathbf{1}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{4a + 4\sqrt{5}}{a^{2} - 5} =\]

\[= \frac{4 \cdot (a + \sqrt{5})}{(a - \sqrt{5})(a + \sqrt{5})} = \frac{4}{a - \sqrt{5}}\]

\[2)\ \frac{\sqrt{28} - 2\sqrt{2a}}{6a - 21} =\]

\[= \frac{\sqrt{2} \cdot \left( \sqrt{14} - 2\sqrt{a} \right)}{3 \cdot (2a - 7)} =\]

\[= \frac{2 \cdot \left( \sqrt{7} - \sqrt{2a} \right)}{3 \cdot (2a - 7)} =\]

\[= \frac{- 2 \cdot \left( \sqrt{7} - \sqrt{2a} \right)}{3 \cdot \left( \sqrt{7} - \sqrt{2a} \right)\left( \sqrt{7} + \sqrt{2a} \right)} =\]

\[= \frac{- 2}{3 \cdot (\sqrt{7} + \sqrt{2a})}\]

\[3)\ \frac{a + 4\sqrt{\text{ab}} + 4b}{a - 4b} =\]

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

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

\[4)\ \frac{x^{2} - 6y}{x^{2} + 6y - x\sqrt{24y}} =\]

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

\[= \frac{x + \sqrt{6y}}{x - \sqrt{6y}}\]

\[5)\ \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a^{3}} + \sqrt{b^{3}}} =\]

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

\[= \frac{1}{a - \sqrt{\text{ab}} + b}\]

\[6)\ \frac{m\sqrt{m} - 27}{\sqrt{m} - 3} = \frac{\sqrt{m^{3}} - 27}{\sqrt{m} - 3} =\]

\[= \frac{\left( \sqrt{m} - 3 \right)\left( m + 3\sqrt{m} + 9 \right)}{\left( \sqrt{m} - 3 \right)} =\]

\[= m + 3\sqrt{m} + 9\]