Lösung 1.2:5a

Aus Online Mathematik Brückenkurs 1

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We begin by calculating the numerator in the main fraction:
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<center> [[Image:1_2_5a.gif]] </center>
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<math>\frac{2}{\frac{1}{7}-\frac{1}{15}}=\frac{2}{\frac{1\centerdot 15}{7\centerdot 15}-\frac{1\centerdot 7}{15\centerdot 7}}=\frac{2}{\frac{15-7}{7\centerdot 15}}=\frac{2}{\frac{8}{7\centerdot 15}}</math>
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Note that we keep
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<math>7\centerdot 15</math>
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as it is, and do not multiply it to give
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<math>105</math>
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, because this will make the task
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of cancellation later simpler. We calculate the fraction on the right-hand side by multiplying top and bottom by
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<math>7\centerdot {15}/{8}\;</math>
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:
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<math>\frac{2}{\frac{8}{7\centerdot 15}}=\frac{2\centerdot \frac{7\centerdot 5}{8}}{\frac{8}{7\centerdot 15}\centerdot \frac{7\centerdot 5}{8}}=\frac{2\centerdot 7\centerdot 15}{8}</math>
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If we now divide up
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<math>8</math>
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and
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<math>15</math>
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into their smallest possible integer factors,
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<math>8=2\centerdot 2\centerdot 2</math>
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and
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<math>15=3\centerdot 5</math>
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, we see that the answer in simplified form will be
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<math>\frac{2\centerdot 7\centerdot 15}{8}=\frac{2\centerdot 7\centerdot 3\centerdot 5}{2\centerdot 2\centerdot 2}=\frac{7\centerdot 3\centerdot 5}{2\centerdot 2}=\frac{105}{4}</math>

Version vom 14:51, 11. Sep. 2008

We begin by calculating the numerator in the main fraction:


\displaystyle \frac{2}{\frac{1}{7}-\frac{1}{15}}=\frac{2}{\frac{1\centerdot 15}{7\centerdot 15}-\frac{1\centerdot 7}{15\centerdot 7}}=\frac{2}{\frac{15-7}{7\centerdot 15}}=\frac{2}{\frac{8}{7\centerdot 15}}


Note that we keep \displaystyle 7\centerdot 15 as it is, and do not multiply it to give \displaystyle 105 , because this will make the task of cancellation later simpler. We calculate the fraction on the right-hand side by multiplying top and bottom by \displaystyle 7\centerdot {15}/{8}\;


\displaystyle \frac{2}{\frac{8}{7\centerdot 15}}=\frac{2\centerdot \frac{7\centerdot 5}{8}}{\frac{8}{7\centerdot 15}\centerdot \frac{7\centerdot 5}{8}}=\frac{2\centerdot 7\centerdot 15}{8}


If we now divide up \displaystyle 8 and \displaystyle 15 into their smallest possible integer factors, \displaystyle 8=2\centerdot 2\centerdot 2 and \displaystyle 15=3\centerdot 5 , we see that the answer in simplified form will be


\displaystyle \frac{2\centerdot 7\centerdot 15}{8}=\frac{2\centerdot 7\centerdot 3\centerdot 5}{2\centerdot 2\centerdot 2}=\frac{7\centerdot 3\centerdot 5}{2\centerdot 2}=\frac{105}{4}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:5a“