Lösung 3.1:5b

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In order to eliminate
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In order to eliminate <math>\sqrt[3]{7} = 7^{1/3}</math> from the denominator, we can multiply the top and bottom of the fraction by <math>7^{2/3}</math>. The denominator becomes <math>7^{1/3}\cdot 7^{2/3} = 7^{1/3+2/3} = 7^1 = 7</math> and we get
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<math>\sqrt[3]{7}=7^{{1}/{3}\;}</math>
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from the denominator, we can multiply the top and bottom of the fraction by
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<math>7^{{2}/{3}\;}</math>. The denominator becomes
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<math>7^{{1}/{3}\;}\centerdot 7^{{2}/{3}\;}=7^{{1}/{3+{2}/{3}\;}\;}=7^{1}=7</math>
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and we get
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{{Displayed math||<math>\frac{1}{\sqrt[3]{7}} = \frac{1}{7^{1/3}} = \frac{1}{7^{1/3}}\cdot \frac{7^{2/3}}{7^{2/3}} = \frac{7^{2/3}}{7}\,\textrm{.}</math>}}
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<math>\frac{1}{\sqrt[3]{7}}=\frac{1}{7^{{1}/{3}\;}}=\frac{1}{7^{{1}/{3}\;}}\centerdot \frac{7^{{2}/{3}\;}}{7^{{2}/{3}\;}}=\frac{7^{{2}/{3}\;}}{7}.</math>
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Version vom 11:22, 30. Sep. 2008

In order to eliminate \displaystyle \sqrt[3]{7} = 7^{1/3} from the denominator, we can multiply the top and bottom of the fraction by \displaystyle 7^{2/3}. The denominator becomes \displaystyle 7^{1/3}\cdot 7^{2/3} = 7^{1/3+2/3} = 7^1 = 7 and we get

Vorlage:Displayed math