Lösung 3.1:5b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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- | In order to eliminate | + | 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 |
- | <math>\sqrt[3]{7}=7^ | + | |
- | from the denominator, we can multiply the top and bottom of the | + | |
- | <math>7^ | + | |
- | <math>7^ | + | |
- | and we get | + | |
- | + | {{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>}} | |
- | <math>\frac{1}{\sqrt[3]{7}}=\frac{1}{7^ | + |
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