frac{sqrt{2}}{sqrt{2 + sqrt{3}} - sqrt{2 - sq | vedclass

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Question

(B) Let $x = \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}$.Multiply the numerator and denominator by $\sqrt{2}$:$x = \frac{\sqrt{2} 	im

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$1$

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(B) Let $x = \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}$.Multiply the numerator and denominator by $\sqrt{2}$:$x = \frac{\sqrt{2} 	imes \sqrt{2}}{\sqrt{2}(\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}})}$$x = \frac{2}{\sqrt{4 + 2\sqrt{3}} - \sqrt{4 - 2\sqrt{3}}}$Note that $4 + 2\sqrt{3} = (\sqrt{3} + 1)^2$ and $4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$.Substituting these values:$x = \frac{2}{(\sqrt{3} + 1) - (\sqrt{3} - 1)}$$x = \frac{2}{\sqrt{3} + 1 - \sqrt{3} + 1}$$x = \frac{2}{2} = 1$.

$\frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}} = $

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