# How do you rationalize the denominator and simplify sqrt22/sqrt33?

$\frac{\sqrt{22}}{\sqrt{33}} = \frac{\sqrt{6}}{3}$
From the given $\frac{\sqrt{22}}{\sqrt{33}} = \frac{\sqrt{22}}{\sqrt{33}} \cdot \frac{\sqrt{33}}{\sqrt{33}}$
$= \frac{\sqrt{22 \cdot 33}}{\sqrt{33}} ^ 2 = \frac{\sqrt{2 \cdot 11 \cdot 3 \cdot 11}}{33} = \frac{\sqrt{6 \cdot {11}^{2}}}{33} = \frac{11 \sqrt{6}}{33} = \frac{\sqrt{6}}{3}$