Question

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

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$\sqrt{2}$

Given, $x=2+\sqrt{3}$ and $xy=1$

$\Rightarrow y=\frac{1}{x}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$

$\sqrt{x}=\sqrt{2+\sqrt{3}}=\frac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}\right)-\frac{\sqrt{3}+1}{2}$

$\sqrt{2}+\sqrt{x}=\sqrt{2}+\frac{\sqrt{2}}{\sqrt{2}}$

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

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

$\sqrt{2}-\sqrt{y}=\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}$

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

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

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

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

$=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}$

$=\frac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-3\sqrt{2}+2\sqrt{6}-3\sqrt{6}-3\sqrt{2}}{(3+\sqrt{3})(3-\sqrt{3})}$

$=\frac{6\sqrt{2}}{9-3}$

$=\sqrt{2}$