Sure! Here's the English translation of your math question: **If** \(x = \frac{\sqrt{3}}{2}\), then what is the value of \[ \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}}? \]

Q.Sure! Here's the English translation of your math question: **If** \(x = \frac{\sqrt{3}}{2}\), then what is the value of \[ \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}}? \] (a) \(2\sqrt3\) (b) \(\cfrac{1}{\sqrt3}\) (c) \(\sqrt3\) (d) \(\sqrt5\)

Answer: C

\[ \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = \frac{(\sqrt{1+x} + \sqrt{1-x})^2}{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})} = \frac{(1+x) + (1-x) + 2\sqrt{1 - x^2}}{(1+x) - (1-x)} = \frac{2 + 2\sqrt{1 - x^2}}{2x} = \frac{1 + \sqrt{1 - x^2}}{x} = \frac{1 + \sqrt{1 - \frac{3}{4}}}{\frac{\sqrt{3}}{2}} = \frac{1 + \sqrt{\frac{1}{4}}}{\frac{\sqrt{3}}{2}} = \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]