How do you simplify $\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}}$ $(sqrt3 -sqrt6)/(sqrt3 +sqrt 6)$ ?

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How do you simplify $\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}}$ $(sqrt3 -sqrt6)/(sqrt3 +sqrt 6)$ ?

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Answer 1 · 2017-07-06T00:45:03

$- 3 + 2 \sqrt{2}$ $-3+2sqrt2$

Explanation:

We can rationalize the denominator by remembering that:

$(\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = a - b$ $(sqrta+sqrtb)(sqrta-sqrtb)=a-b$

In this case:

$\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}}$ $(sqrt3-sqrt6)/(sqrt3+sqrt6)$

$\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}} (1)$ $(sqrt3-sqrt6)/(sqrt3+sqrt6)(1)$

$\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}} (\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} - \sqrt{6}})$ $(sqrt3-sqrt6)/(sqrt3+sqrt6)((sqrt3-sqrt6)/(sqrt3-sqrt6))$

$\frac{{(\sqrt{3} - \sqrt{6})}^{2}}{3 - 6}$ $(sqrt3-sqrt6)^2/(3-6)$

$\frac{3 + 6 - 2 \sqrt{18}}{- 3}$ $(3+6-2sqrt18)/(-3)$

$\frac{9 - 6 \sqrt{2}}{- 3} = - 3 + 2 \sqrt{2}$ $(9-6sqrt2)/(-3)=-3+2sqrt2$

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How do you simplify $\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}}$ $(sqrt3 -sqrt6)/(sqrt3 +sqrt 6)$ ?

1 Answer

Explanation:

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How do you simplify $\frac{\sqrt{3} - \sqrt{6}}{\sqrt{3} + \sqrt{6}}$ $(sqrt3 -sqrt6)/(sqrt3 +sqrt 6)$ ?