Question

The expression $2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}$ equals to

• A. $2$
• B. $2-\sqrt{2}$
• C. $2+\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is A $2$

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

$=2+\sqrt{2}+\frac{1}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}+\frac{1}{\sqrt{2}-2}.\times \frac{\sqrt{2}+2}{\sqrt{2}+2}$ .......... rationalization

$=2+\sqrt{2}+\frac{2-\sqrt{2}}{(4-2)}+\frac{\sqrt{2}+2}{(2-4)}$

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

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