Question #c0fa0

Jul 29, 2017

See a solution process below:

Explanation:

To rationalize the denominator we need to remove all of the radicals from the denominator by multiplying by the appropriate form of $1$. For this type of denominator remember the rule:

$\left(a + b\right) \times \left(a - b\right) = {a}^{2} - {b}^{2}$

$\frac{2 \sqrt{27} + \sqrt{8}}{2 \sqrt{27} + \sqrt{8}} \times \frac{2 \sqrt{6}}{2 \sqrt{27} - \sqrt{8}} \implies$

$\frac{2 \sqrt{6} \left(2 \sqrt{27} + \sqrt{8}\right)}{{\left(2 \sqrt{27}\right)}^{2} - {\left(\sqrt{8}\right)}^{2}} \implies$

$\frac{\left(2 \sqrt{6} \cdot 2 \sqrt{27}\right) + \left(2 \sqrt{6} \cdot \sqrt{8}\right)}{\left(4 \cdot 27\right) - 8} \implies$

$\frac{4 \sqrt{6} \sqrt{27} + 2 \sqrt{6} \sqrt{8}}{108 - 8} \implies$

$\frac{4 \sqrt{6 \cdot 27} + 2 \sqrt{6 \cdot 8}}{100} \implies$

$\frac{4 \sqrt{162} + 2 \sqrt{48}}{100} \implies$

$\frac{4 \sqrt{81 \cdot 2} + 2 \sqrt{16 \cdot 3}}{100} \implies$

$\frac{4 \sqrt{81} \sqrt{2} + 2 \sqrt{16} \sqrt{3}}{100} \implies$

$\frac{\left(4 \cdot 9 \sqrt{2}\right) + \left(2 \cdot 4 \sqrt{3}\right)}{100} \implies$

$\frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \cdot 9 \sqrt{2}\right) + \left(2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \sqrt{3}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{100}}} 25} \implies$

$\frac{9 \sqrt{2} + 2 \sqrt{3}}{25}$