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 $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.First,calculate $x + y$ and

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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 $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.First,calculate $x + y$ and $xy$:$x + y = \frac{(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5} - \sqrt{2})^2}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{(5 + 2 + 2\sqrt{10}) + (5 + 2 - 2\sqrt{10})}{5 - 2} = \frac{14}{3}$.$xy = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} 	imes \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} = 1$.Now,$3x^2 - 3y^2 + 4xy = 3(x - y)(x + y) + 4xy$.We know $(x - y)^2 = (x + y)^2 - 4xy = (\frac{14}{3})^2 - 4(1) = \frac{196}{9} - 4 = \frac{196 - 36}{9} = \frac{160}{9}$.So,$x - y = \frac{\sqrt{160}}{3} = \frac{4\sqrt{10}}{3}$.Substituting these values into the expression:$3(\frac{4\sqrt{10}}{3})(\frac{14}{3}) + 4(1) = \frac{56\sqrt{10}}{3} + 4 = \frac{56\sqrt{10} + 12}{3} = \frac{1}{3}[56\sqrt{10} + 12]$.

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

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