Решебник по алгебре 8 класс Мерзляк ФГОС Задание 1126

\[\boxed{\mathbf{1126}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{\sqrt{x}}{\sqrt{x} - 3} - \frac{x}{x - 9} =\]

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

\[= \frac{x + 3\sqrt{x} - x}{x - 9} = \frac{3\sqrt{x}}{x - 9}\]

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

\[= \frac{\sqrt{\text{bc}} + \sqrt{b} \cdot \left( \sqrt{b} - \sqrt{c} \right)}{\sqrt{c} \cdot \left( \sqrt{b} - \sqrt{c} \right)}\text{\ \ }:\frac{\sqrt{b}}{\sqrt{b} - \sqrt{c}} =\]

\[= \frac{\sqrt{\text{bc}} + b - \sqrt{\text{bc}}}{\sqrt{c} \cdot \left( \sqrt{b} - \sqrt{c} \right)}\ :\frac{\sqrt{b}}{\sqrt{b} - \sqrt{c}} =\]

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