Lösung 1.2:3b

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K (Lösning 1.2:3b moved to Solution 1.2:3b: Robot: moved page)
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If we divide the denominators in succession by
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<center> [[Image:1_2_3b.gif]] </center>
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<math>2</math>
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, we see that
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<math>\begin{align}
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& 24=2\centerdot 2\centerdot 2\centerdot 3 \\
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& 40=2\centerdot 2\centerdot \centerdot 5 \\
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& 16=2\centerdot 2\centerdot 2\centerdot 2 \\
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\end{align}</math>
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i.e. they all have a factor
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<math>2\centerdot 2\centerdot 2=8</math>
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in common,
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<math>\frac{1}{3\centerdot 8}+\frac{1}{5\centerdot 8}-\frac{1}{2\centerdot 8}</math>
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,
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and hence we do not need to take
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<math>8</math>
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as a factor when we multiply the top and bottom of each fraction by the product of the other fractions' denominators, but instead we
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obtain the lowest common denominator by multiplying top and bottom by the other factors,
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<math>2,\ 3</math>
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and
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<math>5</math>
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:
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<math>\frac{1\centerdot 2\centerdot 5}{3\centerdot 8\centerdot 2\centerdot 5}+\frac{1\centerdot 2\centerdot 3}{5\centerdot 8\centerdot 2\centerdot 3}-\frac{1\centerdot 3\centerdot 5}{2\centerdot 8\centerdot 3\centerdot 5}=\frac{10}{240}+\frac{6}{240}-\frac{15}{240}</math>
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The LCD is
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<math>240</math>
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and the answer is
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<math>\frac{10}{240}+\frac{6}{240}-\frac{15}{240}=\frac{10+6-15}{240}=\frac{1}{240}</math>

Version vom 13:10, 11. Sep. 2008

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


\displaystyle \begin{align} & 24=2\centerdot 2\centerdot 2\centerdot 3 \\ & 40=2\centerdot 2\centerdot \centerdot 5 \\ & 16=2\centerdot 2\centerdot 2\centerdot 2 \\ \end{align}


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


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

and hence we do not need to take \displaystyle 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, \displaystyle 2,\ 3 and \displaystyle 5


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


The LCD is \displaystyle 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“