Lösung 3.1:4b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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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 | <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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\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] | \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] | ||
&= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,} | &= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,} | ||
Version vom 08:37, 22. Okt. 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
| \displaystyle \begin{align}
\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] &= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,} \end{align} |
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{.}
