Lösung 3.1:5b

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In order to eliminate
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<center> [[Image:3_1_5b.gif]] </center>
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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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<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>

Version vom 14:34, 22. 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}\;}\centerdot 7^{{2}/{3}\;}=7^{{1}/{3+{2}/{3}\;}\;}=7^{1}=7 and we get


\displaystyle \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}.

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