Lösung 1.3:4e

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Because
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Because <math>5^{9} = 5^{8+1} = 5^{8}\cdot 5^{1} = 5^{8}\cdot 5</math>, the two terms inside the brackets have <math>5^{8}</math> as a common factor and can therefore be taken outside the bracket
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<math>5^{9}=5^{8+1}=5^{8}\centerdot 5^{1}=5^{8}\centerdot 5</math>,
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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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{{Displayed math||<math>\begin{align}
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\bigl(5^{8}+5^{9}\bigr)^{-1} &= \bigl(5^{8}+5^{8}\cdot 5\bigr)^{-1} = \bigl(5^{8}\cdot (1+5)\bigr)^{-1}\\[5pt]
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&= \bigl(5^{8}\cdot 6\bigr)^{-1} = 5^{8\cdot (-1)}\cdot 6^{-1} = 5^{-8}\cdot 6^{-1}.
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\end{align}</math>}}
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<math>\begin{align}
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Furthermore, <math>625 = 5\cdot 125 = 5\cdot 5\cdot 25 = 5\cdot 5\cdot 5\cdot 5 = 5^{4}</math> and we obtain
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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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{{Displayed math||
<math>\begin{align}
<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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625\cdot \bigl(5^{8}+5^{9}\bigr)^{-1} &= 5^{4}\cdot 5^{-8}\cdot 6^{-1} = 5^{4-8}\cdot 6^{-1}\\[5pt]
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& \\
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&= 5^{-4}\cdot 6^{-1} = \frac{1}{5^{4}}\cdot \frac{1}{6}\\[5pt]
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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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&= \frac{1}{5^{4}\cdot 6} = \frac{1}{5\cdot 5\cdot 5\cdot 5\cdot 6}\\[5pt]
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& \\
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&= \frac{1}{3750}\,\textrm{.}
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& =\frac{1}{3750} \\
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\end{align}</math>}}
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\end{align}</math>
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Version vom 14:19, 22. Sep. 2008

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

Vorlage:Displayed math

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

Vorlage:Displayed math

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