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If \(\frac{\sqrt3-1}{\sqrt3+1}\) = x + y √3, find the values of x and y.

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Given, \(\frac{\sqrt3-1}{\sqrt3+1}\) = x + y√3

\(\frac{\sqrt3-1}{\sqrt3+1}\) x \(\frac{\sqrt3-1}{\sqrt3-1}\) = \(\frac{(\sqrt3-1)(\sqrt3-1)}{3-1}\)\(\frac{4-2\sqrt3}{2}\) = 2 - √3

So, x = 2, y = -1

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