Lösung 3.1:5a
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Version vom 14:22, 22. Okt. 2008
If we multiply the top and bottom of the fraction by \displaystyle \sqrt{12}, the new denominator will be \displaystyle \sqrt{12}\cdot\sqrt{12} = 12 and we will get rid of the root sign in the denominator
| \displaystyle \frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.} |
This expression can be simplified even further if we write \displaystyle 12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3 and take \displaystyle 2^2 out from under the root, we get
| \displaystyle \frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.} |
