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यदि x = \(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\) व y = 1/x, तब सिद्ध कीजिये कि x2 + y2 + xy = 99

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  x = \(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\), y = 1/x = \(\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\)

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