Answer: B
\(xy=1\) Or, \(y=\cfrac{1}{x}=\cfrac{\sqrt7-\sqrt3}{\sqrt7+\sqrt3}\) Therefore, \(x+y=\cfrac{\sqrt7+\sqrt3}{\sqrt7-\sqrt3}+\cfrac{\sqrt7-\sqrt3}{\sqrt7+\sqrt3}\) Or, \(x+y\) \(=\cfrac{(\sqrt7+\sqrt3)^2+(\sqrt7-\sqrt3)^2}{(\sqrt7+\sqrt3)(\sqrt7-\sqrt3)}\) \(=\cfrac{2[(\sqrt7)^2+(\sqrt3)^2]}{(\sqrt7)^2-(\sqrt3)^2}\) \(=\cfrac{2(7+3)}{7-3}\) \(=\cfrac{20}{4}=5\) Now, \(\cfrac{x^2+xy+y^2}{x^2-xy+y^2}\) \(= \cfrac{x^2+y^2+xy}{x^2+y^2-xy}\) \(= \cfrac{(x+y)^2-2xy+xy}{(x+y)^2-2xy-xy}\) \(= \cfrac{(x+y)^2-xy}{(x+y)^2-3xy}\) \(= \cfrac{(5)^2-1}{(5)^2-3\times 1}\) \(= \cfrac{25-1}{25-3}\) \(= \cfrac{24}{22}= \cfrac{12}{11}\)
\(xy=1\) Or, \(y=\cfrac{1}{x}=\cfrac{\sqrt7-\sqrt3}{\sqrt7+\sqrt3}\) Therefore, \(x+y=\cfrac{\sqrt7+\sqrt3}{\sqrt7-\sqrt3}+\cfrac{\sqrt7-\sqrt3}{\sqrt7+\sqrt3}\) Or, \(x+y\) \(=\cfrac{(\sqrt7+\sqrt3)^2+(\sqrt7-\sqrt3)^2}{(\sqrt7+\sqrt3)(\sqrt7-\sqrt3)}\) \(=\cfrac{2[(\sqrt7)^2+(\sqrt3)^2]}{(\sqrt7)^2-(\sqrt3)^2}\) \(=\cfrac{2(7+3)}{7-3}\) \(=\cfrac{20}{4}=5\) Now, \(\cfrac{x^2+xy+y^2}{x^2-xy+y^2}\) \(= \cfrac{x^2+y^2+xy}{x^2+y^2-xy}\) \(= \cfrac{(x+y)^2-2xy+xy}{(x+y)^2-2xy-xy}\) \(= \cfrac{(x+y)^2-xy}{(x+y)^2-3xy}\) \(= \cfrac{(5)^2-1}{(5)^2-3\times 1}\) \(= \cfrac{25-1}{25-3}\) \(= \cfrac{24}{22}= \cfrac{12}{11}\)