If x = frac{sqrt{5} + sqrt{2}}{sqrt{5} - sqrt | vedclass

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(B) Given $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.Note that $xy = 1$,so $y = \

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$\frac{1}{3}[56\sqrt{10} + 12]$

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(B) Given $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.Note that $xy = 1$,so $y = \frac{1}{x}$.First,calculate $x + y$:$x + y = \frac{(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5} - \sqrt{2})^2}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{(5 + 2 + 2\sqrt{10}) + (5 + 2 - 2\sqrt{10})}{5 - 2} = \frac{14}{3}$.Next,calculate $x - y$:$x - y = \frac{(\sqrt{5} + \sqrt{2})^2 - (\sqrt{5} - \sqrt{2})^2}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{(7 + 2\sqrt{10}) - (7 - 2\sqrt{10})}{3} = \frac{4\sqrt{10}}{3}$.Now,evaluate $3x^2 - 3y^2 + 4xy = 3(x^2 - y^2) + 4(1) = 3(x - y)(x + y) + 4$.Substituting the values:$3 	imes \left(\frac{4\sqrt{10}}{3}ight) 	imes \left(\frac{14}{3}ight) + 4 = \frac{56\sqrt{10}}{3} + 4 = \frac{56\sqrt{10} + 12}{3} = \frac{1}{3}[56\sqrt{10} + 12]$.

If $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$,then $3x^2 + 4xy - 3y^2 = $

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