Lösung 3.1:5a

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K (Lösning 3.1:5a moved to Solution 3.1:5a: Robot: moved page)
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{{NAVCONTENT_START}}
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If we multiply the top and bottom of the fraction by
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<center> [[Image:3_1_5a.gif]] </center>
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<math>\sqrt{12}</math>, the new denominator will be
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{{NAVCONTENT_STOP}}
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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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<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
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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>

Version vom 14:28, 22. Sep. 2008

If we multiply the top and bottom of the fraction by \displaystyle \sqrt{12}, the new denominator will be \displaystyle \sqrt{12}\centerdot \sqrt{12}=12 and we will get rid of the root sign in the denominator:


\displaystyle \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}


This expression can be simplified even further if we write \displaystyle \text{12}=\text{2}\centerdot \text{6}=\text{2}\centerdot \text{2}\centerdot \text{3}=\text{2}^{\text{2}}\centerdot \text{3 } and take \displaystyle \text{2}^{\text{2}} out from under the root, We get


\displaystyle \frac{\sqrt{12}}{6}=\frac{2\sqrt{3}}{6}=\frac{2\sqrt{3}}{2\centerdot 3}=\frac{\sqrt{3}}{3}.

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