Rationalise

http://img80.imageshack.us/img80/2046/333sco.jpg

Could someone show most of the main steps in achieving the rationalised result of http://img638.imageshack.us/img638/323/24275821.jpg

I tried many methods and failed.

thank you

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- Feb 3rd 2011, 04:27 AMMonster32432421Rationalise
Rationalise

http://img80.imageshack.us/img80/2046/333sco.jpg

Could someone show most of the main steps in achieving the rationalised result of http://img638.imageshack.us/img638/323/24275821.jpg

I tried many methods and failed.

thank you - Feb 3rd 2011, 04:38 AMe^(i*pi)
Did you multiply top and bottom by the conjugate of the denominator?

$\displaystyle \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} $

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

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