\[\boxed{\text{502\ (502).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Вспомним.
Решение.
\[\textbf{а)}\ \frac{x\sqrt{x} - y\sqrt{y}}{\sqrt{x} - \sqrt{y}} = \frac{\sqrt{x^{3}} - \sqrt{y^{3}}}{\sqrt{x} - \sqrt{y}} =\]
\[= x + \sqrt{\text{xy}} + y\]
\[\textbf{б)}\ \frac{\sqrt{a} + \sqrt{b}}{a\sqrt{a} + b\sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a^{3}} + \sqrt{b^{3}}} =\]
\[= \frac{1}{a - \sqrt{\text{ab}} + b}\]
\[\textbf{в)}\ \frac{2\sqrt{2} - x\sqrt{x}}{2 + \sqrt{2x} + x} = \frac{\sqrt{2^{3}} - \sqrt{x^{3}}}{2 + \sqrt{2x} + x} =\]
\[= \frac{\left( \sqrt{2} - \sqrt{x} \right)\left( 2 + \sqrt{2x} + x \right)}{2 + \sqrt{2x} + x} =\]
\[= \sqrt{2} - \sqrt{x}\]
\[\textbf{г)}\ \frac{a - \sqrt{3a} + 3}{a\sqrt{a} + 3\sqrt{3}} = \frac{a - \sqrt{3a} + 3}{\sqrt{a^{3}} + \sqrt{3^{3}}} =\]
\[= \frac{a - \sqrt{3a} + 3}{\left( \sqrt{a} + \sqrt{3} \right)\left( a - \sqrt{3a} + 3 \right)} =\]
\[= \frac{1}{\sqrt{a} + \sqrt{3}}\ \]