10) \(\left(\frac{\sqrt a-\sqrt b}{a\sqrt b+b\sqrt a}+\frac{\sqrt a+\sqrt b}{a\sqrt b-b\sqrt a}\right)\frac{\sqrt{ab}}{a+b}\);

Поскольку \(a\sqrt b+b\sqrt a=\sqrt{ab}(\sqrt a+\sqrt b)\), а \(a\sqrt b-b\sqrt a=\sqrt{ab}(\sqrt a-\sqrt b)\), получаем

\[\left(\frac{\sqrt a-\sqrt b}{\sqrt{ab}(\sqrt a+\sqrt b)}+\frac{\sqrt a+\sqrt b}{\sqrt{ab}(\sqrt a-\sqrt b)}\right)\frac{\sqrt{ab}}{a+b}.\]

\[=\frac1{\sqrt{ab}}\left(\frac{\sqrt a-\sqrt b}{\sqrt a+\sqrt b}+\frac{\sqrt a+\sqrt b}{\sqrt a-\sqrt b}\right)\frac{\sqrt{ab}}{a+b}.\]

Складываем дроби:

\[\frac{(\sqrt a-\sqrt b)^2+(\sqrt a+\sqrt b)^2}{a-b}=\frac{2a+2b}{a-b}.\]

Итак, \[\frac{2(a+b)}{a-b}\cdot\frac1{a+b}=\frac2{a-b}.\]

Ответ: \(\frac2{a-b}\).