Lösung 1.3:5f

Aus Online Mathematik Brückenkurs 1

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The whole expression is quite complicated, so it can be useful to simplify the terms
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<center> [[Image:1_3_5f.gif]] </center>
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<math>\left( 125^{\frac{1}{3}} \right)^{2}</math>
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and
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<math>\left( 27^{\frac{1}{3}} \right)^{-2}</math>
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first:
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<math>\left( 125^{\frac{1}{3}} \right)^{2}=125^{\frac{1}{3}\centerdot 2}=125^{\frac{2}{3}}</math>
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<math>\left( 27^{\frac{1}{3}} \right)^{-2}=27^{\frac{1}{3}\centerdot \left( -2 \right)}=27^{-\frac{2}{3}}</math>
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Then, the bases
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<math>125,\ \ 27</math>
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and
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<math>9</math>
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can be rewritten as
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<math>\begin{align}
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& 125=5\centerdot 25=5\centerdot 5\centerdot 5=5^{3} \\
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& \\
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& 27=3\centerdot 9=3\centerdot 3\centerdot 3=3^{3} \\
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& \\
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& 9=3\centerdot 3=3^{2} \\
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\end{align}</math>
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With the help of the power rules,
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<math>\begin{align}
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& \left( 125^{\frac{1}{3}} \right)^{2}\centerdot \left( 27^{\frac{1}{3}} \right)^{-2}\centerdot 9^{\frac{1}{2}}=125^{\frac{2}{3}}\centerdot 27^{-\frac{2}{3}}\centerdot 9^{\frac{1}{2}} \\
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& \\
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& =\left( 5^{3} \right)^{\frac{2}{3}}\centerdot \left( 3^{3} \right)^{-\frac{2}{3}}\centerdot \left( 3^{2} \right)^{\frac{1}{2}}=5^{3\centerdot \frac{2}{3}}\centerdot 3^{3\centerdot \left( -\frac{2}{3} \right)}\centerdot 3^{2\centerdot \frac{1}{2}} \\
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& \\
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& =5^{2}\centerdot 3^{-2}\centerdot 3^{1}=5^{2}\centerdot 3^{-2+1}=5^{2}\centerdot 3^{-1}=5\centerdot 5\centerdot \frac{1}{3} \\
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& \\
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& \frac{25}{3} \\
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\end{align}</math>

Version vom 12:49, 15. Sep. 2008

The whole expression is quite complicated, so it can be useful to simplify the terms \displaystyle \left( 125^{\frac{1}{3}} \right)^{2} and \displaystyle \left( 27^{\frac{1}{3}} \right)^{-2} first:


\displaystyle \left( 125^{\frac{1}{3}} \right)^{2}=125^{\frac{1}{3}\centerdot 2}=125^{\frac{2}{3}}


\displaystyle \left( 27^{\frac{1}{3}} \right)^{-2}=27^{\frac{1}{3}\centerdot \left( -2 \right)}=27^{-\frac{2}{3}}


Then, the bases \displaystyle 125,\ \ 27 and \displaystyle 9 can be rewritten as


\displaystyle \begin{align} & 125=5\centerdot 25=5\centerdot 5\centerdot 5=5^{3} \\ & \\ & 27=3\centerdot 9=3\centerdot 3\centerdot 3=3^{3} \\ & \\ & 9=3\centerdot 3=3^{2} \\ \end{align}


With the help of the power rules,


\displaystyle \begin{align} & \left( 125^{\frac{1}{3}} \right)^{2}\centerdot \left( 27^{\frac{1}{3}} \right)^{-2}\centerdot 9^{\frac{1}{2}}=125^{\frac{2}{3}}\centerdot 27^{-\frac{2}{3}}\centerdot 9^{\frac{1}{2}} \\ & \\ & =\left( 5^{3} \right)^{\frac{2}{3}}\centerdot \left( 3^{3} \right)^{-\frac{2}{3}}\centerdot \left( 3^{2} \right)^{\frac{1}{2}}=5^{3\centerdot \frac{2}{3}}\centerdot 3^{3\centerdot \left( -\frac{2}{3} \right)}\centerdot 3^{2\centerdot \frac{1}{2}} \\ & \\ & =5^{2}\centerdot 3^{-2}\centerdot 3^{1}=5^{2}\centerdot 3^{-2+1}=5^{2}\centerdot 3^{-1}=5\centerdot 5\centerdot \frac{1}{3} \\ & \\ & \frac{25}{3} \\ \end{align}

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