Rationalise

**Monster32432421** · February 3rd 2011, 04:27 AM

Rationalise

Could someone show most of the main steps in achieving the rationalised result of

I tried many methods and failed.
thank you

**e^(i*pi)** · February 3rd 2011, 04:38 AM

Did you multiply top and bottom by the conjugate of the denominator?

$\dfrac{(2-\sqrt{2})-\sqrt{3}}{(2+\sqrt{2})-\sqrt{3}} \cdot \dfrac{(2+\sqrt{2})+\sqrt{3}}{(2+\sqrt{2})+\sqrt{3 }} = \dfrac{(4-2)+ (2-\sqrt{2})\sqrt{3} - (2+\sqrt{2})\sqrt{3}-3}{4 + 2\sqrt{2}+2-3}$

$= \dfrac{2\sqrt{3}-\sqrt{6}-2\sqrt{3}+\sqrt{6}-1}{3+2\sqrt{2}} = -\dfrac{1}{3+2\sqrt{2}}$

You should be able to multiply by $\dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}$ to rationalise

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