Lösung 1.3:5c

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We can write
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We can write <math>9 = 3\cdot 3 = 3^{2}</math> and then the power rules give
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<math>9=3\centerdot 3=3^{2}</math>
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and then the power rules give
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{{Displayed math||<math>9^{\frac{3}{2}} = \bigl(3^{2}\bigr)^{\frac{3}{2}} = 3^{2\cdot \frac{3}{2}} = 3^{3} = 3\cdot 3\cdot 3 = 27\,</math>.}}
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<math>9^{\frac{3}{2}}=\left( 3^{2} \right)^{\frac{3}{2}}=3^{2\centerdot \frac{3}{2}}=3^{3}=3\centerdot 3\centerdot 3=27</math>
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Version vom 14:26, 22. Sep. 2008

We can write \displaystyle 9 = 3\cdot 3 = 3^{2} and then the power rules give

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:5c“