Question : If $x=\frac{4\sqrt{ab}}{\sqrt a+ \sqrt b}$, then what is the value of $\frac{x+2\sqrt{a}}{x-2\sqrt a}+\frac{x+2\sqrt{b}}{x-2\sqrt b}$(when $a\neq b$)?

Option 1: 0

Option 2: 2

Option 3: 4

Option 4: $\frac{(\sqrt a+\sqrt b)}{(\sqrt a - \sqrt b)}$

Correct Answer: 2

Solution : Given:
$x=\frac{4\sqrt ab}{\sqrt a+\sqrt b}$
Equation $=\frac{x+2\sqrt a}{x-2\sqrt a}+\frac{x+2\sqrt b}{x-2\sqrt b}$
Put the value of $x$ in equation:
$=\frac{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}+2\sqrt{a}}{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-2\sqrt a}+\frac{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}+2\sqrt{b}}{\frac{4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-2\sqrt b}$
$=\frac{\frac{4\sqrt{ab}+2a+2\sqrt {ab}}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2a-2\sqrt {ab}}{\sqrt{a}+\sqrt{b}}}+\frac{\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2\sqrt {ab}-2b}{\sqrt{a}+\sqrt{b}}}$
$=\frac{\frac{4\sqrt{ab}+2a+2\sqrt {a}b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2a-2\sqrt {a}b}{\sqrt{a}+\sqrt{b}}}+\frac{\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{\sqrt{a}+\sqrt{b}}}{\frac{4\sqrt{ab}-2\sqrt {a}b-2b}{\sqrt{a}+\sqrt{b}}}$
$=\frac{4\sqrt{ab}+2a+2\sqrt {a}b}{4\sqrt{ab}-2a-2\sqrt {ab}}+\frac{4\sqrt{ab}+2\sqrt {ab}+2b}{4\sqrt{ab}-2\sqrt {ab}-2b}$
$=\frac{2}{2}\left [\frac{2\sqrt{ab}+a+\sqrt {a}b}{2\sqrt{ab}-a-\sqrt {a}b} \right ]+\frac{2}{2}\left[\frac{2\sqrt{ab}+\sqrt {ab}+b}{2\sqrt{ab}-\sqrt {a}b-b}\right]$
$=\frac{3\sqrt{ab}+a}{\sqrt{ab}-a}+\frac{3\sqrt{ab}+b}{\sqrt{ab}-b}$
$=\frac{\frac{3\sqrt{ab}+a}{\sqrt a}}{\frac{\sqrt{ab}-a}{\sqrt a}}+\frac{\frac{3\sqrt{ab}+b}{\sqrt b}}{\frac{\sqrt{ab}-a}{\sqrt b}}$
$=\frac{3\sqrt b+\sqrt a}{\sqrt b -\sqrt a }+\frac{3\sqrt a+\sqrt b}{\sqrt a -\sqrt b }$
$=\frac{3\sqrt b+\sqrt a}{\sqrt b -\sqrt a }-\frac{3\sqrt a+\sqrt b}{\sqrt b -\sqrt a }$
$=\frac{3\sqrt b+\sqrt a-3\sqrt a -\sqrt b}{\sqrt b-\sqrt a}$
$=\frac{2\sqrt b-2\sqrt a}{\sqrt b-\sqrt a}$
$=\frac{2(\sqrt b-\sqrt a)}{\sqrt b-\sqrt a}$
$=2$
Hence, the correct answer is 2.