Lösung 1.3:1d

Aus Online Mathematik Brückenkurs 1

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By using the power rules, we can rewrite the expression,
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<math>\left( \frac{2}{3} \right)^{-3}=\frac{2^{-3}}{3^{-3}}=\frac{\frac{1}{2^{3}}}{\frac{1}{3^{3}}}=\frac{\frac{1}{2^{3}}\centerdot 3^{3}}{\frac{1}{3^{3}}\centerdot 3^{3}}=\frac{\frac{3^{3}}{2^{3}}}{1}=\frac{3^{3}}{2^{3}}</math>
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and then carry out the calculation:
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<math>\frac{3^{3}}{2^{3}}=\frac{3\centerdot 3\centerdot 3}{2\centerdot 2\centerdot 2}=\frac{27}{8}</math>

Version vom 11:10, 15. Sep. 2008

By using the power rules, we can rewrite the expression,


\displaystyle \left( \frac{2}{3} \right)^{-3}=\frac{2^{-3}}{3^{-3}}=\frac{\frac{1}{2^{3}}}{\frac{1}{3^{3}}}=\frac{\frac{1}{2^{3}}\centerdot 3^{3}}{\frac{1}{3^{3}}\centerdot 3^{3}}=\frac{\frac{3^{3}}{2^{3}}}{1}=\frac{3^{3}}{2^{3}}

and then carry out the calculation:


\displaystyle \frac{3^{3}}{2^{3}}=\frac{3\centerdot 3\centerdot 3}{2\centerdot 2\centerdot 2}=\frac{27}{8}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:1d“