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 <math>\bigl(125^{\frac{1}{3}}\bigr)^{2}</math> and <math>\bigl(27^{\frac{1}{3}}\bigr)^{-2}</math> first, | The whole expression is quite complicated, so it can be useful to simplify the terms <math>\bigl(125^{\frac{1}{3}}\bigr)^{2}</math> and <math>\bigl(27^{\frac{1}{3}}\bigr)^{-2}</math> first, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\bigl(125^{\frac{1}{3}}\bigr)^{2} &= 125^{\frac{1}{3}\cdot 2} = 125^{\frac{2}{3}}\,,\\[5pt] | \bigl(125^{\frac{1}{3}}\bigr)^{2} &= 125^{\frac{1}{3}\cdot 2} = 125^{\frac{2}{3}}\,,\\[5pt] | ||
\bigl(27^{\frac{1}{3}}\bigr)^{-2} &= 27^{\frac{1}{3}\cdot (-2)} = 27^{-\frac{2}{3}}\,\textrm{.}\end{align}</math>}} | \bigl(27^{\frac{1}{3}}\bigr)^{-2} &= 27^{\frac{1}{3}\cdot (-2)} = 27^{-\frac{2}{3}}\,\textrm{.}\end{align}</math>}} | ||
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Then, the bases 125, 27 and 9 can be rewritten as | Then, the bases 125, 27 and 9 can be rewritten as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
125 &= 5\cdot 25 = 5\cdot 5\cdot 5 = 5^{3},\\ | 125 &= 5\cdot 25 = 5\cdot 5\cdot 5 = 5^{3},\\ | ||
27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3},\\ | 27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3},\\ | ||
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With the help of the power rules, | With the help of the power rules, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\bigl(125^{\frac{1}{3}}\bigr)^{2}\cdot\bigl(27^{\frac{1}{3}}\bigr)^{-2}\cdot 9^{\frac{1}{2}} &= 125^{\frac{2}{3}}\cdot 27^{-\frac{2}{3}}\cdot 9^{\frac{1}{2}}\\[5pt] | \bigl(125^{\frac{1}{3}}\bigr)^{2}\cdot\bigl(27^{\frac{1}{3}}\bigr)^{-2}\cdot 9^{\frac{1}{2}} &= 125^{\frac{2}{3}}\cdot 27^{-\frac{2}{3}}\cdot 9^{\frac{1}{2}}\\[5pt] | ||
&= \bigl(5^{3}\bigr)^{\frac{2}{3}}\cdot \bigl(3^{3}\bigr)^{-\frac{2}{3}}\cdot \bigl(3^{2}\bigr)^{\frac{1}{2}}\\[5pt] | &= \bigl(5^{3}\bigr)^{\frac{2}{3}}\cdot \bigl(3^{3}\bigr)^{-\frac{2}{3}}\cdot \bigl(3^{2}\bigr)^{\frac{1}{2}}\\[5pt] | ||
Version vom 08:19, 22. Okt. 2008
The whole expression is quite complicated, so it can be useful to simplify the terms \displaystyle \bigl(125^{\frac{1}{3}}\bigr)^{2} and \displaystyle \bigl(27^{\frac{1}{3}}\bigr)^{-2} first,
| \displaystyle \begin{align}
\bigl(125^{\frac{1}{3}}\bigr)^{2} &= 125^{\frac{1}{3}\cdot 2} = 125^{\frac{2}{3}}\,,\\[5pt] \bigl(27^{\frac{1}{3}}\bigr)^{-2} &= 27^{\frac{1}{3}\cdot (-2)} = 27^{-\frac{2}{3}}\,\textrm{.}\end{align} |
Then, the bases 125, 27 and 9 can be rewritten as
| \displaystyle \begin{align}
125 &= 5\cdot 25 = 5\cdot 5\cdot 5 = 5^{3},\\ 27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3},\\ 9 &= 3\cdot 3 = 3^{2}\textrm{.} \end{align} |
With the help of the power rules,
| \displaystyle \begin{align}
\bigl(125^{\frac{1}{3}}\bigr)^{2}\cdot\bigl(27^{\frac{1}{3}}\bigr)^{-2}\cdot 9^{\frac{1}{2}} &= 125^{\frac{2}{3}}\cdot 27^{-\frac{2}{3}}\cdot 9^{\frac{1}{2}}\\[5pt] &= \bigl(5^{3}\bigr)^{\frac{2}{3}}\cdot \bigl(3^{3}\bigr)^{-\frac{2}{3}}\cdot \bigl(3^{2}\bigr)^{\frac{1}{2}}\\[5pt] &= 5^{3\cdot\frac{2}{3}}\cdot 3^{3\cdot (-\frac{2}{3})}\cdot 3^{2\cdot\frac{1}{2}}\\[5pt] &= 5^{2}\cdot 3^{-2}\cdot 3^{1}\\[5pt] &= 5^{2}\cdot 3^{-2+1}\\[5pt] &= 5^{2}\cdot 3^{-1}\\[5pt] &= 5\cdot 5\cdot \frac{1}{3}\\[5pt] &= \frac{25}{3}\,\textrm{.} \end{align} |
