$x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} a n d Y = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} t h e n t h e v a l u e o f \frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}}$

dsuc.created: 2 years ago | dsuc.updated: 1 year ago

dsuc.updated: 1 year ago

Given, $\frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}}$

$∴ x + y = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} + \frac{\sqrt{5} - 1}{\sqrt{5} + 1} = \frac{{(\sqrt{5} + 1)}^{2} + {(\sqrt{5} - 1)}^{2}}{(\sqrt{5} - 1) (\sqrt{5} + 1)} = \frac{{(\sqrt{5})}^{2} + 2 \times \sqrt{5} \times 1 + 1^{2} + {(\sqrt{5})}^{2} - 2 \times \sqrt{5} \times 1 + 1^{2}}{{(\sqrt{5})}^{2} - (1)} = \frac{5 + 2 \sqrt{5} + 1 + 5 - 2 \sqrt{5} + 1}{5 - 1} = \frac{12}{4} = 3 A n d, x y = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} + 1} = 1 N o w, x^{2} + x y + y^{2} = x^{2} + 2 x y + y^{2} - x y = {(x + y)}^{2} - x y = 3^{2} - 1 = 9 - 1 = 8 A n d, x^{2} - x y + y^{2} = x^{2} + 2 x y + y^{2} - 3 x y = {(x + y)}^{2} - 3 x y = 3^{2} - (3 \times 1) = 9 - 3 = 6 ∴ \frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}} = \frac{8}{6} = \frac{4}{3}$