Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
$(\frac{1}{a}+\frac{1}{b}-\frac{2c}{ab})(a+b+2c):(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}-\frac{4c^2}{a^2{b}^2})$

Answer 1

$(\frac{1}{a}+\frac{1}{b}-\frac{2c}{ab})(a+b+2c)=\frac{(a+b-2c)(a+b+2c)}{ab}=\frac{(a+b{)}^2-4c^2}{ab}$

$(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}-\frac{4c^2}{a^2{b}^2})=\frac{a^2+b^2+2ab-4c^2}{a^2{b}^2}=\frac{(a+b{)}^2-4c^2}{a^2{b}^2}$

$⟹(\frac{1}{a}+\frac{1}{b}-\frac{2c}{ab})(a+b+2c):(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}-\frac{4c^2}{a^2{b}^2})=1:\frac{1}{ab}=ab$