Lösung 1.3:4e

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{{NAVCONTENT_START}}
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Because
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<center> [[Image:1_3_4e.gif]] </center>
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<math>5^{9}=5^{8+1}=5^{8}\centerdot 5^{1}=5^{8}\centerdot 5</math>,
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{{NAVCONTENT_STOP}}
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the two terms inside the brackets have
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<math>5^{8}</math>
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as a common factor
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and can therefore be taken outside the bracket.
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<math>\begin{align}
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& \left( 5^{8}+5^{9} \right)^{-1}=\left( 5^{8}+5^{8}\centerdot 5 \right)^{-1}=\left( 5^{8}\centerdot \left( 1+5 \right) \right)^{-1} \\
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& \\
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& =\left( 5^{8}\centerdot 6 \right)^{-1}=5^{8\centerdot \left( -1 \right)}\centerdot 6^{-1}=5^{-8}\centerdot 6^{-1}. \\
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\end{align}</math>
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Furthermore,
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<math>625=5\centerdot 125=5\centerdot 5\centerdot 25=5\centerdot 5\centerdot 5\centerdot 5=5^{4}</math>
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and we obtain
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<math>\begin{align}
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& 625\centerdot \left( 5^{8}+5^{9} \right)^{-1}=5^{4}\centerdot 5^{-8}\centerdot 6^{-1}=5^{4-8}\centerdot 6^{-1} \\
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& \\
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& =5^{-4}\centerdot 6^{-1}=\frac{1}{5^{4}}\centerdot \frac{1}{6}=\frac{1}{5^{4}\centerdot 6}=\frac{1}{5\centerdot 5\centerdot 5\centerdot 5\centerdot 6} \\
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& \\
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& =\frac{1}{3750} \\
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\end{align}</math>

Version vom 11:57, 15. Sep. 2008

Because \displaystyle 5^{9}=5^{8+1}=5^{8}\centerdot 5^{1}=5^{8}\centerdot 5, the two terms inside the brackets have \displaystyle 5^{8} as a common factor and can therefore be taken outside the bracket.


\displaystyle \begin{align} & \left( 5^{8}+5^{9} \right)^{-1}=\left( 5^{8}+5^{8}\centerdot 5 \right)^{-1}=\left( 5^{8}\centerdot \left( 1+5 \right) \right)^{-1} \\ & \\ & =\left( 5^{8}\centerdot 6 \right)^{-1}=5^{8\centerdot \left( -1 \right)}\centerdot 6^{-1}=5^{-8}\centerdot 6^{-1}. \\ \end{align}

Furthermore, \displaystyle 625=5\centerdot 125=5\centerdot 5\centerdot 25=5\centerdot 5\centerdot 5\centerdot 5=5^{4} and we obtain


\displaystyle \begin{align} & 625\centerdot \left( 5^{8}+5^{9} \right)^{-1}=5^{4}\centerdot 5^{-8}\centerdot 6^{-1}=5^{4-8}\centerdot 6^{-1} \\ & \\ & =5^{-4}\centerdot 6^{-1}=\frac{1}{5^{4}}\centerdot \frac{1}{6}=\frac{1}{5^{4}\centerdot 6}=\frac{1}{5\centerdot 5\centerdot 5\centerdot 5\centerdot 6} \\ & \\ & =\frac{1}{3750} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.3:4e“