Simplify: $\;$ $\dfrac{2a \sqrt{1 + x^2}}{x + \sqrt{1 + x^2}}$ $\;$ when $\;$ $x = \dfrac{1}{2} \left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right)$, $\;\;$ $a > 0, \; b > 0$
Given: $\;\;\;$ $x = \dfrac{1}{2} \left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right) = \dfrac{a - b}{2 \sqrt{ab}}$
Value of $\;$ $\sqrt{1 + x^2}$ $\;$ when $\;$ $x = \dfrac{1}{2} \left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right)$ $\;$ is
$\sqrt{1 + \left(\dfrac{a - b}{2ab}\right)^2}$
$= \sqrt{1 + \dfrac{a^2 - 2ab + b^2}{4ab}}$
$= \sqrt{\dfrac{4ab + a^2 - 2ab + b^2}{4ab}}$
$= \sqrt{\dfrac{a^2 + 2ab + b^2}{4ab}}$
$= \sqrt{\dfrac{\left(a + b\right)^2}{4ab}}$
$= \dfrac{a + b}{2 \sqrt{ab}}$
Value of $\;$ $x + \sqrt{1 + x^2}$ $\;$ when $\;$ $x = \dfrac{1}{2} \left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right)$ $\;$ is
$\dfrac{a - b}{2 \sqrt{ab}} + \dfrac{a + b}{2 \sqrt{ab}}$
$= \dfrac{2a}{2 \sqrt{ab}}$
$= \dfrac{a}{\sqrt{ab}}$
$\therefore \;$ Value of given expression $\;$ $\dfrac{2a \sqrt{1 + x^2}}{x + \sqrt{1 + x^2}}$ $\;$ when $\;$ $x = \dfrac{1}{2} \left(\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}}\right)$ $\;$ is
$\dfrac{2a \left(\dfrac{a + b}{2 \sqrt{ab}}\right)}{\dfrac{a}{\sqrt{ab}}}$
$= a + b$