Решебник по алгебре 8 класс Мерзляк ФГОС Задание 573

 

\[\boxed{\mathbf{573\ (573).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\sqrt{2} \cdot \sqrt{2 + \sqrt{2}} \cdot \sqrt{4 - 2 - \sqrt{2}} = 2\]

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

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

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

\[2 = 2.\]

\[\boxed{\mathbf{5}\mathbf{7}\mathbf{3}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\frac{x - 25}{\sqrt{x} - 5} = \frac{\left( \sqrt{x} - 5 \right)\left( \sqrt{x} + 5 \right)}{\left( \sqrt{x} - 5 \right)} =\]

\[= \sqrt{x} + 5\]

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

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

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

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

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

\[4)\ \frac{\sqrt{10} + \sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5} \cdot \left( \sqrt{2} + 1 \right)}{\sqrt{5}} =\]

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

\[5)\ \frac{23 - \sqrt{23}}{\sqrt{23}} =\]

\[= \frac{\sqrt{23} \cdot \left( \sqrt{23} - 1 \right)}{\sqrt{23}} = \sqrt{23} - 1\]

\[6)\ \frac{\sqrt{24} - \sqrt{28}}{\sqrt{54} - \sqrt{63}} = \frac{2\sqrt{6} - 2\sqrt{7}}{3\sqrt{6} - 3\sqrt{7}} =\]

\[= \frac{2 \cdot \left( \sqrt{6} - \sqrt{7} \right)}{3 \cdot \left( \sqrt{6} - \sqrt{7} \right)} = \frac{2}{3}\]

\[7)\ \frac{\sqrt{a} - \sqrt{b}}{a - 2\sqrt{\text{ab}} + b} =\]

\[= \frac{\left( \sqrt{a} - \sqrt{b} \right)}{\left( \sqrt{a} - \sqrt{b} \right)^{2}} = \frac{1}{\sqrt{a} - \sqrt{b}}\]

\[8)\ \frac{b - 8\sqrt{b} + 16}{\sqrt{b} - 4} = \frac{\left( \sqrt{b} - 4 \right)^{2}}{\left( \sqrt{b} - 4 \right)} =\]

\[= \sqrt{b} - 4\]