Lösung 3.1:5a

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If we multiply the top and bottom of the fraction by
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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
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<math>\sqrt{12}</math>, the new denominator will be
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<math>\sqrt{12}\centerdot \sqrt{12}=12</math>
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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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<math>\frac{2}{\sqrt{12}}=\frac{2}{\sqrt{12}}\centerdot \frac{\sqrt{12}}{\sqrt{12}}=\frac{2\sqrt{12}}{12}=\frac{2\sqrt{12}}{2\centerdot 6}=\frac{\sqrt{12}}{6}</math>
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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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This expression can be simplified even further if we write
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<math>\text{12}=\text{2}\centerdot \text{6}=\text{2}\centerdot \text{2}\centerdot \text{3}=\text{2}^{\text{2}}\centerdot \text{3 }</math>
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and take
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<math>\text{2}^{\text{2}}</math>
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out from under the root, We get
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<math>\frac{\sqrt{12}}{6}=\frac{2\sqrt{3}}{6}=\frac{2\sqrt{3}}{2\centerdot 3}=\frac{\sqrt{3}}{3}.</math>
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Version vom 11:18, 30. Sep. 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

Vorlage:Displayed math

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

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:5a“