Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
$(\sqrt{\sqrt{3}+\sqrt{2}}-(\sqrt{3}-\sqrt{2}{)}^{1/2}\left)\right((\sqrt{3}+\sqrt{2}{)}^{1/2}+\sqrt{\sqrt{3}-\sqrt{2}}{)}^{-1}-\cos \frac{5\pi }{4}$

Answer 1

Let $a=\sqrt{3},b=\sqrt{2}$

$\therefore$ The given question becomes

$\left(\frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}\right)-\cos \frac{5\pi }{4}$

now rationalizing the numerator by multiplying and dividing the denominator

$\Rightarrow \left(\frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}\right)\left(\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}\right)-\cos \frac{5\pi }{4}$

$\Rightarrow \left[\frac{(a+b)-(a-b)}{{(\sqrt{a+b}+\sqrt{a-b})}^2}\right]+\frac{1}{\sqrt{2}}$

$\Rightarrow =\left[\frac{2b}{(2a+2\sqrt{a^2-b^2})}\right]+\frac{1}{\sqrt{2}}$

We know that $a^2-b^2=1$

On substituting the values of a,b we get

$\Rightarrow =\left[\frac{\sqrt{2}}{(\sqrt{3}+1)}\right]+\frac{1}{\sqrt{2}}$

Now again rationalizing the denominator we get

$\Rightarrow \frac{\sqrt{2}(\sqrt{3}-1)}{2}+\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}+\sqrt{2}}{2}=\sqrt{\frac{3}{2}}$