Lösung 1.3:1d

Aus Online Mathematik Brückenkurs 1

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By using the power rules, we can rewrite the expression,
By using the power rules, we can rewrite the expression,
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{{Displayed math||<math>\left( \frac{2}{3} \right)^{-3} = \frac{2^{-3}}{3^{-3}} = \frac{\,\dfrac{1}{2^{3}}\,}{\,\dfrac{1}{3^{3}}\,} = \frac{\,\dfrac{1}{2^{3}}\cdot 3^{3}\,}{\,\dfrac{1}{\rlap{\,/}3^{3}}\cdot {}\rlap{\,/}3^{3}\,} = \frac{\,\dfrac{3^{3}}{2^{3}}\,}{1} = \frac{3^{3}}{2^{3}}\,,</math>}}
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{{Abgesetzte Formel||<math>\left( \frac{2}{3} \right)^{-3} = \frac{2^{-3}}{3^{-3}} = \frac{\,\dfrac{1}{2^{3}}\,}{\,\dfrac{1}{3^{3}}\,} = \frac{\,\dfrac{1}{2^{3}}\cdot 3^{3}\,}{\,\dfrac{1}{\rlap{\,/}3^{3}}\cdot {}\rlap{\,/}3^{3}\,} = \frac{\,\dfrac{3^{3}}{2^{3}}\,}{1} = \frac{3^{3}}{2^{3}}\,,</math>}}
and then carry out the calculation
and then carry out the calculation
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{{Displayed math||<math>\frac{3^{3}}{2^{3}} = \frac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2} = \frac{27}{8}\,</math>.}}
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{{Abgesetzte Formel||<math>\frac{3^{3}}{2^{3}} = \frac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2} = \frac{27}{8}\,</math>.}}

Version vom 08:16, 22. Okt. 2008

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

\displaystyle \left( \frac{2}{3} \right)^{-3} = \frac{2^{-3}}{3^{-3}} = \frac{\,\dfrac{1}{2^{3}}\,}{\,\dfrac{1}{3^{3}}\,} = \frac{\,\dfrac{1}{2^{3}}\cdot 3^{3}\,}{\,\dfrac{1}{\rlap{\,/}3^{3}}\cdot {}\rlap{\,/}3^{3}\,} = \frac{\,\dfrac{3^{3}}{2^{3}}\,}{1} = \frac{3^{3}}{2^{3}}\,,

and then carry out the calculation

\displaystyle \frac{3^{3}}{2^{3}} = \frac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2} = \frac{27}{8}\,.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:1d“