Lösung 1.2:3b

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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If we divide the denominators in succession by 2, we see that
If we divide the denominators in succession by 2, we see that
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
24&=2\cdot 2\cdot 2\cdot 3\,,\\
24&=2\cdot 2\cdot 2\cdot 3\,,\\
40&=2\cdot 2\cdot 2\cdot 5\,,\\
40&=2\cdot 2\cdot 2\cdot 5\,,\\
Zeile 9: Zeile 9:
i.e. they all have a factor <math>2\cdot 2\cdot 2=8</math> in common,
i.e. they all have a factor <math>2\cdot 2\cdot 2=8</math> in common,
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{{Displayed math||<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}
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{{Abgesetzte Formel||<math>\frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,</math>.}}
Hence we do not need to take 8 as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we
Hence we do not need to take 8 as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we
obtain the lowest common denominator by multiplying top and bottom by the other factors, 2, 3 and 5,
obtain the lowest common denominator by multiplying top and bottom by the other factors, 2, 3 and 5,
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{{Displayed math||<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}
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{{Abgesetzte Formel||<math>\frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,</math>.}}
The LCD is 240 and the answer is
The LCD is 240 and the answer is
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{{Displayed math||<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}
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{{Abgesetzte Formel||<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,</math>.}}

Version vom 08:14, 22. Okt. 2008

If we divide the denominators in succession by 2, we see that

\displaystyle \begin{align} 
 24&=2\cdot 2\cdot 2\cdot 3\,,\\ 
 40&=2\cdot 2\cdot 2\cdot 5\,,\\ 
 16&=2\cdot 2\cdot 2\cdot 2\,,\\

\end{align}

i.e. they all have a factor \displaystyle 2\cdot 2\cdot 2=8 in common,

\displaystyle \frac{1}{3\cdot 8}+\frac{1}{5\cdot 8}-\frac{1}{2\cdot 8}\,.

Hence we do not need to take 8 as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we obtain the lowest common denominator by multiplying top and bottom by the other factors, 2, 3 and 5,

\displaystyle \frac{1\cdot 2\cdot 5}{3\cdot 8\cdot 2\cdot 5}+\frac{1\cdot 2\cdot 3}{5\cdot 8\cdot 2\cdot 3}-\frac{1\cdot 3\cdot 5}{2\cdot 8\cdot 3\cdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}\,.

The LCD is 240 and the answer is

\displaystyle \frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}\,.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:3b“