Given: \(xy = 1\) So, \(y = \frac{1}{x} = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}\) Now, \[ x + y = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} + \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}} \] \[ = \frac{(\sqrt{7} + \sqrt{3})^2 + (\sqrt{7} - \sqrt{3})^2}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})} = \frac{2[(\sqrt{7})^2 + (\sqrt{3})^2]}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{2(7 + 3)}{7 - 3} = \frac{20}{4} = 5 \] Now, \[ \frac{x^2 + 3xy + y^2}{x^2 - 3xy + y^2} = \frac{x^2 + y^2 + 3xy}{x^2 + y^2 - 3xy} = \frac{(x + y)^2 - 2xy + 3xy}{(x + y)^2 - 2xy - 3xy} = \frac{(x + y)^2 + xy}{(x + y)^2 - 5xy} \] Substitute \(x + y = 5\) and \(xy = 1\): \[ = \frac{5^2 + 1}{5^2 - 5 \times 1} = \frac{25 + 1}{25 - 5} = \frac{26}{20} = \frac{13}{10} \]