Question

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

Answer 1

$\frac{x+2}{x-2}-\frac{x+2\sqrt{2}}{x-2\sqrt{2}}$ (Apply compounds and dividends)
$\frac{a}{b}=\frac{a+b}{a-b}$
So,
$\frac{x+2+(x-2)}{x+2-(x-2)}-\frac{(x+2\sqrt{2})+(x-2\sqrt{2})}{(x+2\sqrt{2})-(x-2\sqrt{2})}$
$\frac{2x}{4}-\frac{2x}{4\sqrt{2}}$
$\frac{2x}{4}(1-\frac{1}{\sqrt{2}})$
$\frac{1}{2}x\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
Now, substitute $x-\frac{4\sqrt{2}}{\sqrt{2}+1}$
$\frac{1}{2}\frac{\text{/}4\sqrt{2}}{\sqrt{2}+1}\frac{\sqrt{2}-1}{\text{/}\sqrt{2}}$
$2\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)$ (Rationalise the denominator)
$2\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}=2\left(\frac{(\sqrt{2}-1{)}^2}{1}\right)$
$=2(2+1-2\sqrt{2})$
$=2(2-2\sqrt{2})$
$=6-4\sqrt{2}$