Question

if x=√5+1/√5-1 and y=√5-1/√5+1 then the value of x^2-xy+y^2/x^2-xy+y^2 =?​

Answer 1

Given,

$\boxed{x = \dfrac{ \sqrt{5} + 1 }{ \sqrt{5} - 1}} \:$

Rationalising the denominator,

$x = \frac{ \sqrt{5} + 1 }{ \sqrt{5} - 1 } \times \frac{ \sqrt{5} + 1 }{ \sqrt{5} + 1 } \\ \\ x = \frac{ {( \sqrt{5} + 1) }^{2} }{( \sqrt{5}) ^{2} - {1}^{2} } \\ \\ x = \frac{5 + 1 + 2 \sqrt{5} }{4} \\ \\ x = \frac{6 + 2 \sqrt{5} }{4} \\ \\ x = \frac{3 + \sqrt{5} }{2}$

$\boxed{y= \dfrac{ \sqrt{5} - 1 }{ \sqrt{5} + 1 }} \:$

Rationalising the denominator,

$y= \frac{ \sqrt{5} - 1 }{ \sqrt{5} + 1 } \times \frac{ \sqrt{5} - 1 }{ \sqrt{5} - 1 } \\ \\ y = \frac{ {( \sqrt{5} - 1) }^{2} }{( \sqrt{5}) ^{2} - {1}^{2} } \\ \\ y = \frac{5 + 1 - 2 \sqrt{5} }{4} \\ \\ y= \frac{6 - 2 \sqrt{5} }{4} \\ \\ y= \frac{3 - \sqrt{5} }{2}$

Now, The question seems to have some mistake as the Numerator and Denominator are same, it's equal to 1.

Correct question :

$\frac{ {x}^{2} - xy + {y}^{2} }{ {x}^{2} + xy + {y}^{2} }$

Solving the expression :

$\frac{ {x}^{2} - 2xy + {y}^{2} + xy }{ {x}^{2} + 2xy + {y}^{2} - xy } \\ \\ = \frac{(x - y) ^{2} + xy}{(x + y) ^{2} - xy}$

Now,

$x - y = \frac{3 + \sqrt{5} }{2} - \frac{3 - \sqrt{5} }{2} = \sqrt{5}$

$x + y = \frac{3 + \sqrt{5} }{2} + \frac{3 - \sqrt{5} }{2} = 3$

$x y = \frac{3 + \sqrt{5} }{2} \times \frac{3 - \sqrt{5} }{2} = \frac{9 - 5}{2} = \frac{4}{2} = 2$

$\frac{ {x}^{2} - xy + {y}^{2} }{ {x}^{2} + xy + {y}^{2} } \\ \\ = \frac{(x - y) ^{2} + xy}{(x + y) ^{2} - xy} \\ \\ = \frac{( \sqrt{5}) ^{2} + 2 }{ {( \sqrt{5}) }^{2} - 2} \\ \\ = \frac{5 + 2}{5 - 2} \\ \\ = \frac{7}{3}$

Therefore, The required answer is 7/3.