Question

If x=2+root 3 , xy=1 then find x/2-x+y/2-y

Answer 1

x=2+√3
xy=1
so,y=1/2+√3
=>y=2-√3

x/2-x+y/2-y
=(2+√3)/2-(2+√3+2-√3)/2+(2-√3)
=1+√3/2-2+2-√3
=2-√3/2(ans)

please mark as brainliest answer if you get help from this

Answer 2

Answer:

-2

Step-by-step explanation:

Here,

x = 2 + √3,

And, xy = 1 ⇒ y = 1/x

$=\frac{1}{2+\sqrt{3}}$

By rationalizing,

$y=\frac{2-\sqrt{3}}{4-3}$

$\implies y=2-\sqrt{3}$

Thus,

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

$=\frac{2+\sqrt{3}}{2-(2+\sqrt{3})} + \frac{2-\sqrt{3}}{2-(2-\sqrt{3})}$

$=\frac{2+\sqrt{3}}{2-2-\sqrt{3})} + \frac{2-\sqrt{3}}{2-2+\sqrt{3}}$

$=\frac{2+\sqrt{3}}{-\sqrt{3})} + \frac{2-\sqrt{3}}{\sqrt{3}}$

$=\frac{-2-\sqrt{3}}{\sqrt{3})} + \frac{2-\sqrt{3}}{\sqrt{3}}$

$=\frac{-2-\sqrt{3}+2-\sqrt{3}}{\sqrt{3}}$

$=\frac{-2\sqrt{3}}{\sqrt{3}}$

$=-2$