Question

If $x=2+\sqrt{3},xy=1,$ then find $\frac{x}{2-x}+\frac{y}{2-y}$

• A. $-3$
• B. $2\sqrt{3}$
• C. $-2$
• D. $\frac{4}{\sqrt{3}}$

Answer 1

The correct option is C

$-2$

Given:$x=2+\sqrt{3},xy=1\frac{x}{2-x}+\frac{y}{2-y}=\frac{x}{2-x}+\frac{xy}{2x-xy}=\frac{x}{2-x}+\frac{1}{2x-1}=\frac{2+\sqrt{3}}{2-(2+\sqrt{3})}+\frac{1}{2(2+\sqrt{3})-1}=\frac{(2+\sqrt{3})}{-\sqrt{3}}+\frac{1}{2(2+\sqrt{3})-1}=\frac{(2+\sqrt{3})}{-\sqrt{3}}+\frac{1}{3+2\sqrt{3}}=\frac{-6-7\sqrt{3}-6+\sqrt{3}}{\sqrt{3}(3+2\sqrt{3})}=\frac{-(12+6\sqrt{3})}{\sqrt{3}(3+2\sqrt{3})}=\frac{-6(2+\sqrt{3})}{3(2+\sqrt{3})}=\frac{-6}{3}=-2.$