\[\boxed{\text{503\ (503).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Разложим\ подкоренные\ \]
\[выражения\ на\ множителе\ и\ \]
\[вынесем\ общий\ множитель\ \]
\[за\ скобки.\]
\[\textbf{а)}\ \frac{\sqrt{70} - \sqrt{30}}{\sqrt{35} - \sqrt{15}} =\]
\[= \frac{\sqrt{7} \cdot \sqrt{10} - \sqrt{3} \cdot \sqrt{10}}{\sqrt{7} \cdot \sqrt{5} - \sqrt{3} \cdot \sqrt{5}} =\]
\[= \frac{\sqrt{10} \cdot \left( \sqrt{7} - \sqrt{3} \right)}{\sqrt{5} \cdot \left( \sqrt{7} - \sqrt{3} \right)} = \frac{\sqrt{2} \cdot \sqrt{5}}{\sqrt{5}} =\]
\[= \sqrt{2}\ \]
\[\textbf{б)}\ \frac{\sqrt{15} - 5}{\sqrt{6} - \sqrt{10}} = \frac{\sqrt{5}\sqrt{3} - \sqrt{5}\sqrt{5}}{\sqrt{3}\sqrt{2} - \sqrt{5}\sqrt{2}} =\]
\[= \frac{\sqrt{5} \cdot \left( \sqrt{3} - \sqrt{5} \right)}{\sqrt{2} \cdot \left( \sqrt{3} - \sqrt{5} \right)} = \frac{\sqrt{5}}{\sqrt{2}}\ \]
\[\textbf{в)}\ \frac{2\sqrt{10} - 5}{4 - \sqrt{10}} = \frac{2\sqrt{5}\sqrt{2} - \sqrt{5}\sqrt{5}}{\sqrt{2}\sqrt{2^{3}} - \sqrt{2}\sqrt{5}} =\]
\[= \frac{\sqrt{5} \cdot \left( 2\sqrt{2} - \sqrt{5} \right)}{\sqrt{2} \cdot \left( 2\sqrt{2} - \sqrt{5} \right)} = \frac{\sqrt{5}}{\sqrt{2}}\ \]
\[\textbf{г)}\ \frac{9 - 2\sqrt{3}}{3\sqrt{6} - 2\sqrt{2}} = \frac{\sqrt{3}\sqrt{3^{3}} - 2\sqrt{3}}{3\sqrt{2}\sqrt{3} - 2\sqrt{2}} =\]
\[= \frac{\sqrt{3} \cdot \left( 3\sqrt{3} - 2 \right)}{\sqrt{2} \cdot \left( 3\sqrt{3} - 2 \right)} = \frac{\sqrt{3}}{\sqrt{2}}\ \]
\[\textbf{д)}\ \frac{2\sqrt{3} + 3\sqrt{2} - \sqrt{6}}{2 + \sqrt{6} - \sqrt{2}} =\]
\[= \frac{\sqrt{2}\sqrt{2}\sqrt{3} + \sqrt{3}\sqrt{3}\sqrt{2} - \sqrt{2}\sqrt{3}}{\sqrt{2}\sqrt{2} + \sqrt{2}\sqrt{3} - \sqrt{2}} =\]
\[= \frac{\sqrt{2}\sqrt{3} \cdot \left( \sqrt{2} + \sqrt{3} - 1 \right)}{\sqrt{2} \cdot \left( \sqrt{2} + \sqrt{3} - 1 \right)} =\]
\[= \sqrt{3}\]
\[\textbf{е)}\ \frac{\left( \sqrt{10} - 1 \right)^{2} - 3}{\sqrt{10} + \sqrt{3} - 1} =\]
\[= \frac{\left( \sqrt{10} - 1 - \sqrt{3} \right)\left( \sqrt{10} - 1 + \sqrt{3} \right)}{\sqrt{10} + \sqrt{3} - 1} =\]
\[= \sqrt{10} - 1 - \sqrt{3}\ \]