If x = frac{4 sqrt{ab}}{sqrt{a} + sqrt{b}},th | vedclass

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(B) Given $x = \frac{4 \sqrt{ab}}{\sqrt{a} + \sqrt{b}}$.We need to find the value of $\frac{x + 2 \sqrt{a}}{x - 2 \sqrt{a}} + \frac{x + 2 \sqrt{b}}{x

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$2$

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(B) Given $x = \frac{4 \sqrt{ab}}{\sqrt{a} + \sqrt{b}}$.We need to find the value of $\frac{x + 2 \sqrt{a}}{x - 2 \sqrt{a}} + \frac{x + 2 \sqrt{b}}{x - 2 \sqrt{b}}$.First,consider the term $\frac{x + 2 \sqrt{a}}{x - 2 \sqrt{a}}$.Substitute $x = \frac{4 \sqrt{ab}}{\sqrt{a} + \sqrt{b}}$:$\frac{\frac{4 \sqrt{ab}}{\sqrt{a} + \sqrt{b}} + 2 \sqrt{a}}{\frac{4 \sqrt{ab}}{\sqrt{a} + \sqrt{b}} - 2 \sqrt{a}} = \frac{4 \sqrt{ab} + 2 \sqrt{a}(\sqrt{a} + \sqrt{b})}{4 \sqrt{ab} - 2 \sqrt{a}(\sqrt{a} + \sqrt{b})} = \frac{4 \sqrt{ab} + 2a + 2 \sqrt{ab}}{4 \sqrt{ab} - 2a - 2 \sqrt{ab}} = \frac{2a + 6 \sqrt{ab}}{2 \sqrt{ab} - 2a} = \frac{a + 3 \sqrt{ab}}{\sqrt{ab} - a} = \frac{\sqrt{a}(\sqrt{a} + 3 \sqrt{b})}{\sqrt{a}(\sqrt{b} - \sqrt{a})} = \frac{\sqrt{a} + 3 \sqrt{b}}{\sqrt{b} - \sqrt{a}}$.Similarly,for $\frac{x + 2 \sqrt{b}}{x - 2 \sqrt{b}}$:$\frac{\frac{4 \sqrt{ab}}{\sqrt{a} + \sqrt{b}} + 2 \sqrt{b}}{\frac{4 \sqrt{ab}}{\sqrt{a} + \sqrt{b}} - 2 \sqrt{b}} = \frac{4 \sqrt{ab} + 2 \sqrt{b}(\sqrt{a} + \sqrt{b})}{4 \sqrt{ab} - 2 \sqrt{b}(\sqrt{a} + \sqrt{b})} = \frac{4 \sqrt{ab} + 2 \sqrt{ab} + 2b}{4 \sqrt{ab} - 2 \sqrt{ab} - 2b} = \frac{6 \sqrt{ab} + 2b}{2 \sqrt{ab} - 2b} = \frac{3 \sqrt{ab} + b}{\sqrt{ab} - b} = \frac{\sqrt{b}(3 \sqrt{a} + \sqrt{b})}{\sqrt{b}(\sqrt{a} - \sqrt{b})} = \frac{3 \sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = -\frac{3 \sqrt{a} + \sqrt{b}}{\sqrt{b} - \sqrt{a}}$.Adding the two results:$\frac{\sqrt{a} + 3 \sqrt{b}}{\sqrt{b} - \sqrt{a}} - \frac{3 \sqrt{a} + \sqrt{b}}{\sqrt{b} - \sqrt{a}} = \frac{\sqrt{a} + 3 \sqrt{b} - 3 \sqrt{a} - \sqrt{b}}{\sqrt{b} - \sqrt{a}} = \frac{2 \sqrt{b} - 2 \sqrt{a}}{\sqrt{b} - \sqrt{a}} = \frac{2(\sqrt{b} - \sqrt{a})}{\sqrt{b} - \sqrt{a}} = 2$.

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 $a \neq b$)?

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