Ta có
$A = 2 \dfrac{\sqrt{x^2-1}}{x - \sqrt{x^2-1}} = 2 \dfrac{\sqrt{x^2-1}(x + \sqrt{x^2-1})}{x^2 - (x^2-1)}$
$= 2 x\sqrt{x^2-1} + x^2-1$
Lại có
$x = \dfrac{1}{2} \left( \dfrac{\sqrt{a}}{\sqrt{b}} + \dfrac{\sqrt{b}}{\sqrt{a}} \right)$
$= \dfrac{a + b}{2\sqrt{ab}} $
Do đó
$x^2- 1 = \dfrac{(a+b)^2}{4ab} - 1 = \dfrac{(a-b)^2}{4ab}$
Vậy
$A = \dfrac{a+b}{2\sqrt{ab}} . \sqrt{\dfrac{(a-b)^2}{4ab}} + \dfrac{(a-b)^2}{4ab}$
$= \dfrac{a+b}{2\sqrt{ab}} . \dfrac{a-b}{2\sqrt{ab}} + \dfrac{(a-b)^2}{4ab}$
$= \dfrac{a^2 - b^2}{4ab} + \dfrac{a^2 + b^2 - 2ab}{4ab}$
$= \dfrac{2a^2 - 2ab}{4ab} = \dfrac{a - b}{2b}$
