Question

X = (√5+1)/(√5-1)
Y = (√5-1)/(√5+1)

then, What is the value of
(x²+xy+y²)/(x²-xy+y²)​

Answer 1

Answer:

$\frac{4}{3}$

Step-by-step explanation:

x=(√5+1)^2/4=(3+√5)/2

y=(√5-1)^2/4=(3-√5)/2

$\left \{ {{x+y=3} \atop {xy=1}} \right.$

x²+xy+y²=(x+y)²-xy=9-1=8

x²-xy+y²=(x+y)²-3xy=9-3=6

THE ANSWER IS $\frac{4}{3}$

Answer 2

The value of $\frac{x^2+xy+y^2}{x^2-xy+y^2}=\frac{4}{3}$

Step-by-step explanation:

$x=\frac{\sqrt 5+1}{\sqrt 5-1}$

$y=\frac{\sqrt 5-1}{\sqrt 5+1}$

By rationalization

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

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

By using identity:$(a+b)^2=a^2+b^2+2ab,(a+b)(a-b)=a^2-b^2$

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

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

$\frac{x^2+xy+y^2}{x^2-xy+y^2}=\frac{\frac{9}{4}+\frac{5}{4}+3\sqrt 5+1+\frac{9}{4}+\frac{5}{4}-3\sqrt 5}{\frac{9}{4}+\frac{5}{4}+3\sqrt 5-1+\frac{9}{4}+\frac{5}{4}-3\sqrt 5}$

$\frac{x^2+xy+y^2}{x^2-xy+y^2}=\frac{4}{3}$

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