Simplify: $\;$ $\left[\left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right) : \left(\sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{b}{a}} - 2\right)\right] : \left(1 + \sqrt{\dfrac{b}{a}}\right)$, $\;\;$ $a > 0, \; b> 0$
The given expression
$\left[\left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right) : \left(\sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{b}{a}} - 2\right)\right] : \left(1 + \sqrt{\dfrac{b}{a}}\right)$ $\;\;\; \cdots \; (1)$
Consider $\;\;\;$ $\left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right) : \left(\sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{b}{a}} - 2\right)$
$= \dfrac{a - b}{\sqrt{ab}} : \dfrac{a + b - 2 \sqrt{ab}}{\sqrt{ab}}$
$= \dfrac{a - b}{\left(\sqrt{a} - \sqrt{b}\right)^2}$
$= \dfrac{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}{\left(\sqrt{a} - \sqrt{b}\right)^2}$
$= \dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of $(2)$, expression $(1)$ becomes
$\dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} : 1 + \sqrt{\dfrac{b}{a}}$
$= \dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} \times \dfrac{\sqrt{a}}{\sqrt{a} + \sqrt{b}}$
$= \dfrac{\sqrt{a}}{\sqrt{a} - \sqrt{b}}$