Answer: 2√2 - 3

Explanation:

The expession written properly is:

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

To rationalize that kind of expressions, this is to eliminate the radicals on the denominator you use conjugate rationalization.

That is, you have to multiply both numerator and denominator times the conjugate of the denominator.

The conjugate of √3+√6 is √3 - √6, so let's do it:

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

To help you with the solution of that expression, I will show each part.

1) Numerator: (√3 - √6) . (√3 - √6) = (√3 - √6)^2 = (√3)^2 - 2√3√6 + (√6)^2 =

= 3 - 2√18 + 6 = 9 - 6√2.

2) Denominator: (√3 + √6).(√3 - √6) = (√3)^2 - (√6)^2 = 3 - 6 = - 3

3) Then the resulting expression is:

9 - 6√2
-----------
-3

Which can be further simplified, dividing by - 3

-3 + 2√2

Answer: 2√2 - 3

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