Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
Simplify the following expressions.
$a^2(1-a^2{)}^{-1/2}:(\frac{1}{1+\left(a\right(1-a^2{)}^{-1/2}{)}^2}\cdot \frac{(1-a^2{)}^{1/2}+a^2(1-a^2{)}^{-1/2}}{1-a^2})\cdot$

Answer 1

$a^2(1-a^2{)}^{\frac{-1}{2}}:(\frac{1}{1+\left(a\right(1-a^2{)}^{\frac{-1}{2}}{)}^2}\cdot \frac{(1-a^2{)}^{\frac{1}{2}}+a^2(1-a^2{)}^{\frac{-1}{2}}}{1-a^2})$

$=\frac{a^2}{\sqrt{1-a^2}}:\frac{1}{1+\frac{a^2}{1-a^2}}\cdot \frac{\sqrt{1-a^2}+\frac{a^2}{\sqrt{1-a^2}}}{1-a^2}$

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

$=\frac{a^2}{\sqrt{1-a^2}}:\frac{1}{\sqrt{1-a^2}}$

$=\frac{a^2}{\sqrt{1-a^2}}\times \sqrt{1-a^2}=a^2$