Lösung 1.2:5a

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We begin by calculating the numerator in the main fraction:
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We begin by calculating the numerator in the main fraction,
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{{Displayed math||<math>\frac{2}{\,\dfrac{1}{7}-\dfrac{1}{15}\vphantom{\Biggl(}\,} = \frac{2}{\,\dfrac{1\cdot 15}{7\cdot 15}-\dfrac{1\cdot 7}{15\cdot 7}\vphantom{\Biggl(}\,} = \frac{2}{\,\dfrac{15-7}{7\cdot 15}\vphantom{\Biggl(}\,} = \frac{2}{\,\dfrac{8}{7\cdot 15}\vphantom{\Biggl(}\,}\,</math>.}}
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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 <math>7\cdot 15</math> as it is, and do not multiply it to give 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 <math>7\cdot 15/8</math>,
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{{Displayed math||<math>\frac{2}{\,\dfrac{8}{7\cdot 15}\vphantom{\Biggl(}\,} = \frac{\,2\cdot \dfrac{7\cdot 15}{8}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}8}{\rlap{/}7\cdot{}\rlap{\,/}15}\cdot \dfrac{\rlap{/}7\cdot{}\rlap{\,/}15}{\rlap{/}8}\vphantom{\Biggl(}\,} = \frac{2\cdot 7\cdot 15}{8}\,</math>.}}
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Note that we keep
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If we now divide up 8 and 15 into their smallest possible integer factors <math>8=2\cdot 2\cdot 2</math> and <math>15=3\cdot 5</math>, we see that the answer in simplified form will be
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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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{{Displayed math||<math>\frac{2\cdot 7\cdot 15}{8}=\frac{\rlap{/}2\cdot 7\cdot 3\cdot 5}{\rlap{/}2\cdot 2\cdot 2}=\frac{7\cdot 3\cdot 5}{2\cdot 2}=\frac{105}{4}\,</math>.}}
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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>
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Version vom 11:24, 19. Sep. 2008

We begin by calculating the numerator in the main fraction,

Vorlage:Displayed math

Note that we keep \displaystyle 7\cdot 15 as it is, and do not multiply it to give 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\cdot 15/8,

Vorlage:Displayed math

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

Vorlage:Displayed math

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