\[\boxed{\text{505\ (505).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \frac{x - \sqrt{\text{xy}} + y}{\sqrt{x} - \sqrt{y}} =\]
\[= \frac{\left( x - \sqrt{\text{xy}} + y \right) \cdot \left( \sqrt{x} + \sqrt{y} \right)}{\left( \sqrt{x} - \sqrt{y} \right) \cdot \left( \sqrt{x} + \sqrt{y} \right)} =\]
\[= \frac{\sqrt{x^{3}} + \sqrt{y^{3}}}{x - y} = \frac{x\sqrt{x} + y\sqrt{y}}{x - y}\ \]
\[\textbf{б)}\ \frac{9 + 3\sqrt{a} + a}{3 + \sqrt{a}} =\]
\[= \frac{\left( 9 + 3\sqrt{a} + a \right) \cdot \left( 3 - \sqrt{a} \right)}{\left( 3 + \sqrt{a} \right) \cdot \left( 3 - \sqrt{a} \right)} =\]
\[= \frac{3^{3} - \left( \sqrt{a} \right)^{3}}{9 - a} = \frac{27 - a\sqrt{a}}{9 - a}\]
\[\textbf{в)}\ \frac{1 - 2\sqrt{x} + 4x}{1 - 2\sqrt{x}} =\]
\[= \frac{\left( 1 - 2\sqrt{x} + 4x \right) \cdot \left( 1 + 2\sqrt{x} \right)}{\left( 1 - 2\sqrt{x} \right) \cdot \left( 1 + 2\sqrt{x} \right)} =\]
\[= \frac{1 + \left( 2\sqrt{x} \right)^{3}}{1 - 4x} = \frac{1 + 8x\sqrt{x}}{1 - 4x}\]
\[\textbf{г)}\ \frac{a^{2}b + 2a\sqrt{b} + 4}{a\sqrt{b} + 2} =\]
\[= \frac{\left( a^{2}b + 2a\sqrt{b} + 4 \right) \cdot \left( a\sqrt{b} - 2 \right)}{\left( a\sqrt{b} + 2 \right) \cdot \left( a\sqrt{b} - 2 \right)} =\]
\[= \frac{\left( a\sqrt{b} \right)^{3} - 2^{3}}{a^{2}b - 4} = \frac{a^{3}b\sqrt{b} - 8}{a^{2}b - 4}\ \]