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

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Solution

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

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

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

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