Lösung 3.1:5a

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If we multiply the top and bottom of the fraction by <math>\sqrt{12}</math>, the new denominator will be <math>\sqrt{12}\cdot\sqrt{12} = 12</math> and we will get rid of the root sign in the denominator
If we multiply the top and bottom of the fraction by <math>\sqrt{12}</math>, the new denominator will be <math>\sqrt{12}\cdot\sqrt{12} = 12</math> and we will get rid of the root sign in the denominator
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{{Displayed math||<math>\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{.}</math>}}
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{{Abgesetzte Formel||<math>\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{.}</math>}}
This expression can be simplified even further if we write <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> and take <math>2^2</math> out from under the root, we get
This expression can be simplified even further if we write <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> and take <math>2^2</math> out from under the root, we get
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{{Displayed math||<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}}

Version vom 08:38, 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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:5a“