Lösung 3.1:4b

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By writing
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By writing <math>0\textrm{.}027</math> as <math>27\cdot 10^{-3}</math>, where
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<math>0.0\text{27 }</math>
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<math>27 = 3\cdot 3\cdot 3 = 3^3</math> and <math>10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3</math> we see that
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as
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<math>\text{27}\cdot \text{1}0^{-\text{3}}</math>, where
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<math>\text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}}</math>
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and
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<math>10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3}</math>
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we see that
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{{Displayed math||<math>\begin{align}
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\sqrt[3]{0\textrm{.}027} &= \sqrt[3]{27\cdot 10^{-3}} = \sqrt[3]{27}\cdot\sqrt[3]{10^{-3}} = \sqrt[3]{3^{3}}\cdot\sqrt[3]{0\textrm{.}1^3}\\[5pt]
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&= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,}
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\end{align}</math>}}
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<math>\begin{align}
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where we have used <math>\sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}</math>
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& \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\
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& =3\centerdot 0.1=0.3 \\
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\end{align}</math>
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where we have used
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<math>\sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a</math>
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Version vom 10:54, 30. Sep. 2008

By writing \displaystyle 0\textrm{.}027 as \displaystyle 27\cdot 10^{-3}, where \displaystyle 27 = 3\cdot 3\cdot 3 = 3^3 and \displaystyle 10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3 we see that

Vorlage:Displayed math

where we have used \displaystyle \sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:4b“