# How do you simplify (3sqrt3)/(-2+sqrt6)?

Jul 21, 2017

See a solution process below:

#### Explanation:

The first step is to rationalize the denominator by multiplying the fraction by the appropriate form of $1$:

$\frac{3 \sqrt{3}}{- 2 + \sqrt{6}} \times \frac{- 2 - \sqrt{6}}{- 2 - \sqrt{6}} \implies$

$\frac{3 \sqrt{3} \left(- 2 - \sqrt{6}\right)}{{\left(- 2\right)}^{2} + \left(- 2 \cdot - \sqrt{6}\right) + \left(- 2 \cdot \sqrt{6}\right) - {\left(\sqrt{6}\right)}^{2}} \implies$

$\frac{\left(- 2 \cdot 3 \sqrt{3}\right) - \left(\sqrt{6} \cdot 3 \sqrt{3}\right)}{4 + 0 - 6} \implies$

$\frac{- 6 \sqrt{3} - 3 \sqrt{6 \cdot 3}}{- 2} \implies$

$\frac{- 6 \sqrt{3}}{- 2} - \frac{3 \sqrt{18}}{- 2} \implies$

$3 \sqrt{3} + \frac{3 \sqrt{9 \cdot 2}}{2} \implies$

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

$3 \sqrt{3} + \frac{3 \cdot 3 \sqrt{2}}{2} \implies$

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

Jul 21, 2017

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

#### Explanation:

$\frac{3 \sqrt{3}}{- 2 + \sqrt{6}}$

$\therefore - \frac{3 \sqrt{3}}{- 2 + \sqrt{6}} \times \frac{- 2 - \sqrt{6}}{- 2 - \sqrt{6}}$

$\therefore = \frac{- 6 \sqrt{3} - 9 \sqrt{2}}{-} 2$

$\therefore = \frac{{\cancel{- 6}}^{3} \sqrt{3}}{\cancel{- 2}} ^ 1 - \frac{9 \sqrt{2}}{-} 2$

$\therefore = 3 \sqrt{3} - \frac{9 \sqrt{2}}{- 2}$

$\therefore = 3 \sqrt{3} + \left(\frac{- 9 \sqrt{2}}{1} \times \frac{1}{-} 2\right)$

$\therefore = 3 \sqrt{3} + \frac{9 \sqrt{2}}{2}$

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$c h e c k$

$\therefore \frac{3 \sqrt{3}}{- 2 + \sqrt{6}} = 11.56011345$

$\therefore \frac{- 6 \sqrt{3} - 9 \sqrt{2}}{-} 2 = 11.56011345$

$\therefore 3 \sqrt{3} + \frac{9 \sqrt{2}}{2} = 11.56011345$