If x = (2+√3)/(2-√3), y = (2-√3)/(2+√3). Then find out the value of (x²+y²-xy)/(x²+y²+xy)?
x = (2+√3)/(2-√3) y = (2-√3)/(2+√3) So x+y = (2+√3)/(2-√3) + (2-√3)/(2+√3) = ((2+√3)²+ (2-√3)²)/((2)²- (√3)²) = 2(4+3)/(4-3) {As (a+b)² + (a-b)² = 2(a^2+b^2 )} = x+y = 14 = xy = 1 = (x^2+y^2-xy)/(x^2+y^2+xy) = ((x+y)²-3xy)/((x+y)²-xy) = ((14)^2- 3)/((14)^2-1) = (196-3)/(196 - 1) = 193/195
