Question

If $x=\frac{4\sqrt{2}}{\sqrt{2}+1}$ then find the value of $\frac{1}{\sqrt{2}}(\frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x-2\sqrt{2}})$

• A. $\sqrt{2}$
• B. $12+8\sqrt{2}/5$
• C. $12-8\sqrt{2}$
• D. $\frac{16\sqrt{2}+24}{5}$

Answer 1

The correct option is A $\sqrt{2}$

$x=\frac{4\sqrt{2}}{\sqrt{2+1}}$
$\frac{1}{\sqrt{2}}(\frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x-2\sqrt{2}})$
Put the value of x
$\frac{1}{\sqrt{2}}(\frac{\frac{4\sqrt{2}}{\sqrt{2+1}}+2}{\frac{4\sqrt{2}}{\sqrt{2+1}}-2}+\frac{\frac{4\sqrt{2}}{\sqrt{2+1}}+2\sqrt{2}}{\frac{4\sqrt{2}}{\sqrt{2+1}}-2\sqrt{2}})$
=$\frac{1}{\sqrt{2}}\left(\frac{4\sqrt{2}+2\sqrt{2}+2}{4\sqrt{2}-2\sqrt{2}-2}\right)+\left(\frac{4\sqrt{2}+4+2\sqrt{2}}{4\sqrt{2}-4-2\sqrt{2}}\right)$
=$\frac{6\sqrt{2}+2}{2\sqrt{2-2}}+\frac{6\sqrt{2}+4}{2\sqrt{2}-4}$
=$\frac{1}{\sqrt{2}}\left(\frac{32-24\sqrt{2}}{16-12\sqrt{2}}\right)$
=$\sqrt{2}$