If x = frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}} | vedclass

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(C) Given $x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ and $y = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$.First,calculate $xy = \left(\frac{\s

Vedclass · Accepted Answer

$\frac{63}{61}$

Vedclass · Answer

(C) Given $x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ and $y = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$.First,calculate $xy = \left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right) \left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\right) = 1$.Next,calculate $x+y = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} + \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{(\sqrt{5}-\sqrt{3})^2 + (\sqrt{5}+\sqrt{3})^2}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{(5+3-2\sqrt{15}) + (5+3+2\sqrt{15})}{5-3} = \frac{16}{2} = 8$.Calculate $x-y = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} - \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{(\sqrt{5}-\sqrt{3})^2 - (\sqrt{5}+\sqrt{3})^2}{5-3} = \frac{(8-2\sqrt{15}) - (8+2\sqrt{15})}{2} = \frac{-4\sqrt{15}}{2} = -2\sqrt{15}$.We need to find $\frac{x^2+xy+y^2}{x^2-xy+y^2}$. This can be rewritten as $\frac{(x+y)^2 - xy}{(x-y)^2 + xy}$.Substituting the values: $\frac{8^2 - 1}{(-2\sqrt{15})^2 + 1} = \frac{64-1}{60+1} = \frac{63}{61}$.

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

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