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Lösung 1.3:4e

Aus Online Mathematik Brückenkurs 1

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K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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
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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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\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]
\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]
&= \bigl(5^{8}\cdot 6\bigr)^{-1} = 5^{8\cdot (-1)}\cdot 6^{-1} = 5^{-8}\cdot 6^{-1}.
&= \bigl(5^{8}\cdot 6\bigr)^{-1} = 5^{8\cdot (-1)}\cdot 6^{-1} = 5^{-8}\cdot 6^{-1}.
Zeile 8: Zeile 8:
Furthermore, <math>625 = 5\cdot 125 = 5\cdot 5\cdot 25 = 5\cdot 5\cdot 5\cdot 5 = 5^{4}</math> and we obtain
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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{{Displayed math||
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{{Abgesetzte Formel||
<math>\begin{align}
<math>\begin{align}
625\cdot \bigl(5^{8}+5^{9}\bigr)^{-1} &= 5^{4}\cdot 5^{-8}\cdot 6^{-1} = 5^{4-8}\cdot 6^{-1}\\[5pt]
625\cdot \bigl(5^{8}+5^{9}\bigr)^{-1} &= 5^{4}\cdot 5^{-8}\cdot 6^{-1} = 5^{4-8}\cdot 6^{-1}\\[5pt]

Version vom 08:18, 22. Okt. 2008

Because 59=58+1=5851=585, the two terms inside the brackets have 58 as a common factor and can therefore be taken outside the bracket

58+59−1=58+585−1=58(1+5)−1=586−1=58(−1)6−1=5−86−1

Furthermore, 625=5125=5525=5555=54 and we obtain

62558+59−1=545−86−1=54−86−1=5−46−1=15461=1546=155556=13750.

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