Question

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

Answer 1

We will simplify the given expression $(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2$ as shown below:

$(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2=\left(\frac{a\sqrt{a}+b\sqrt{b}-\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}\right){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2=\left(\frac{a\sqrt{a}+b\sqrt{b}-\sqrt{a^{2}b}-\sqrt{a{b}^2}}{\sqrt{a}+\sqrt{b}}\right){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2=\left(\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}\right){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2$

$=\left(\frac{a(\sqrt{a}-\sqrt{b})-b(\sqrt{a}-\sqrt{b})}{\sqrt{a}+\sqrt{b}}\right){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2=\left(\frac{(a-b)(\sqrt{a}-\sqrt{b})}{\sqrt{a}+\sqrt{b}}\right){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2=\left(\frac{(a-b)(\sqrt{a}-\sqrt{b})}{\sqrt{a}+\sqrt{b}}\right){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=(\sqrt{a}{)}^2-(\sqrt{b}{)}^2=a-b(\because (x+y\left)\right(x-y)=x^2-y^2)$

Hence, $(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}){\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)}^2=a-b$