Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
$(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}):(\frac{3}{\sqrt{1-a^2}}+1)$

Answer 1

$(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}):(\frac{3}{\sqrt{1-a^2}}+1)$

$=\left(\frac{3+\sqrt{1-a}\sqrt{1+a}}{\sqrt{1+a}}\right)\times \frac{\sqrt{1-a^2}}{3+\sqrt{1-a^2}}$

$=\frac{(3+\sqrt{1-a^2})}{\sqrt{1+a}}\times \frac{\sqrt{1-a}\times \sqrt{1+a}}{(3+\sqrt{1-a^2})}$

$=\sqrt{1-a}$