\[\boxed{\text{506\ (506).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \frac{\sqrt{x} - \sqrt{y}}{\sqrt{x}} =\]
\[= \frac{\left( \sqrt{x} - \sqrt{y} \right) \cdot \left( \sqrt{x} + \sqrt{y} \right)}{\sqrt{x} \cdot \left( \sqrt{x} + \sqrt{y} \right)} =\]
\[= \frac{x - y}{\sqrt{x} \cdot \left( \sqrt{x} + \sqrt{y} \right)} = \frac{x - y}{x + \sqrt{x}y}\]
\[\textbf{б)}\ \frac{a + \sqrt{b}}{a\sqrt{b}} =\]
\[= \frac{\left( a + \sqrt{b} \right) \cdot \left( a - \sqrt{b} \right)}{a\sqrt{b} \cdot \left( a - \sqrt{b} \right)} =\]
\[= \frac{a^{2} - b}{a\sqrt{b} \cdot \left( a - \sqrt{b} \right)} = \frac{a^{2} - b}{a^{2}\sqrt{b} - ab}\]
\[\textbf{в)}\ \frac{7 - \sqrt{a}}{49 - 7\sqrt{a} + a} =\]
\[= \frac{7 - \sqrt{a} \cdot \left( 7 + \sqrt{a} \right)}{49 - 7\sqrt{a} + a \cdot \left( 7 + \sqrt{a} \right)} =\]
\[= \frac{49 - a}{7^{3} - \left( \sqrt{a} \right)^{3}} = \frac{49 - a}{343 + a\sqrt{a}}\]
\[\textbf{г)}\ \frac{\sqrt{\text{mn}} + 1}{mn + \sqrt{\text{mn}} + 1} =\]
\[= \frac{\sqrt{\text{mn}} + 1 \cdot \left( \sqrt{\text{mn}} - 1 \right)}{mn + \sqrt{\text{mn}} + 1 \cdot \left( \sqrt{\text{mn}} - 1 \right)} =\]
\[= \frac{mn - 1}{{\sqrt{\text{mn}}}^{3} - 1} = \frac{mn - 1}{\text{mn}\sqrt{\text{mn}} - 1}\]