Решебник по алгебре 8 класс Мерзляк ФГОС Задание 561

 

\[\boxed{\mathbf{561\ (561).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

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

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

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

\[= \frac{6}{m - 2\sqrt{m}}\]

\[3)\ \frac{\sqrt{y} + 4}{\sqrt{\text{xy}} + y} - \frac{\sqrt{x} - 4}{x + \sqrt{\text{xy}}} =\]

\[= \frac{4x + 4y + 8\sqrt{\text{xy}}}{\sqrt{\text{xy}} \cdot \left( x + y + 2\sqrt{\text{xy}} \right)} =\]

\[= \frac{4 \cdot (x + y + 2\sqrt{\text{xy}})}{\sqrt{\text{xy}} \cdot \left( x + y + 2\sqrt{\text{xy}} \right)} = \frac{4}{\sqrt{\text{xy}}}\]

\[4)\ \frac{\sqrt{a}}{\sqrt{a} + 4} - \frac{a}{a - 16} =\]

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

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

\[= \frac{- 4\sqrt{a}}{a - 16} = \frac{4\sqrt{a}}{16 - a}\]

\[5)\ \frac{a}{\sqrt{\text{ab}} - b} + \frac{\sqrt{b}}{\sqrt{b} - \sqrt{a}} =\]

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

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

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

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

\[6)\ \frac{a + \sqrt{a}}{\sqrt{b}} \cdot \frac{b}{2\sqrt{a} + 2} =\]

\[= \frac{\sqrt{a} \cdot \left( \sqrt{a} + 1 \right) \cdot b}{\sqrt{b} \cdot 2 \cdot (\sqrt{a} + 1)} = \frac{\sqrt{\text{ab}}}{2}\]

\[7)\ \frac{\sqrt{c} - 5}{\sqrt{c}}\ :\frac{c - 25}{3c} =\]

\[= \frac{\left( \sqrt{c} - 5 \right) \cdot 3c}{\left( \sqrt{c} - 5 \right)\left( \sqrt{c} + 5 \right) \cdot \sqrt{c}} =\]

\[= \frac{3\sqrt{c}}{\sqrt{c} + 5}\]

\[8)\ \left( \sqrt{a} - \frac{a}{\sqrt{a} + 1} \right)\ :\frac{\sqrt{a}}{a - 1} =\]

\[= \frac{a + \sqrt{a} - a}{\sqrt{a} + 1} \cdot \frac{a - 1}{\sqrt{a}} =\]

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

\[= \frac{(\sqrt{a} - 1)(\sqrt{a} + 1)}{(\sqrt{a} + 1)} = \sqrt{a} - 1\]

\[9)\ \left( \frac{\sqrt{a} + \sqrt{b}}{\sqrt{b}} + \frac{\sqrt{b}}{\sqrt{a} - \sqrt{b}} \right)\ :\frac{\sqrt{a}}{\sqrt{b}} =\]

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

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

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

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

\[= \sqrt{x}\]

\[\boxed{\mathbf{5}\mathbf{6}\mathbf{1}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 4\sqrt{700} - 27\sqrt{7} =\]

\[= 40\sqrt{7} - 27\sqrt{7} = 13\sqrt{7};\]

\[2)\ \sqrt{75} - 6\sqrt{3} = 5\sqrt{3} - 6\sqrt{3} =\]

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

\[3)\ 2\sqrt{50} - 8\sqrt{2} = 10\sqrt{2} - 8\sqrt{2} =\]

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

\[4)\ 5\sqrt{12} - 7\sqrt{3} = 10\sqrt{3} - 7\sqrt{3} =\]

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

\[5)\ 3\sqrt{72} - 4\sqrt{2} + 2\sqrt{98} =\]

\[= 18\sqrt{2} - 4\sqrt{2} + 14\sqrt{2} = 28\sqrt{2};\]

\[6)\ \frac{1}{3}\sqrt{108} + \sqrt{363} - \frac{2}{9} \cdot \sqrt{243} =\]

\[= 2\sqrt{3} + 11\sqrt{3} - 2\sqrt{3} = 11\sqrt{3}.\]