Lösung 1.2:2c
Aus Online Mathematik Brückenkurs 1
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We divide up the two numerators into the smallest possible integer factors, | We divide up the two numerators into the smallest possible integer factors, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\ | 12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\ | ||
14 &= 2\cdot 7\,\textrm{.} \\ | 14 &= 2\cdot 7\,\textrm{.} \\ | ||
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The expression can thus be written as | The expression can thus be written as | ||
- | {{ | + | {{Abgesetzte Formel|| |
<math>\frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\,</math>.}} | <math>\frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\,</math>.}} | ||
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i.e. we leave out the common factor 2, so that the fractions have the lowest common denominator <math>2\cdot 2\cdot 3\cdot 7</math>, | i.e. we leave out the common factor 2, so that the fractions have the lowest common denominator <math>2\cdot 2\cdot 3\cdot 7</math>, | ||
- | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{1}{12}-\frac{1}{14} &= \frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\\[5pt] | \frac{1}{12}-\frac{1}{14} &= \frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\\[5pt] | ||
&= \frac{1}{2\cdot 2\cdot 3}\cdot \frac{7}{7}-\frac{1}{2\cdot 7}\cdot \frac{2\cdot 3}{2\cdot 3}\\[5pt] | &= \frac{1}{2\cdot 2\cdot 3}\cdot \frac{7}{7}-\frac{1}{2\cdot 7}\cdot \frac{2\cdot 3}{2\cdot 3}\\[5pt] |
Version vom 08:14, 22. Okt. 2008
We divide up the two numerators into the smallest possible integer factors,
\displaystyle \begin{align}
12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\ 14 &= 2\cdot 7\,\textrm{.} \\ \end{align} |
The expression can thus be written as
\displaystyle \frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\,. |
Here, we see that the denominators have a factor 2 in common. We multiply the top and bottom of the first fraction by 7 and the second by \displaystyle 2\cdot 3 i.e. we leave out the common factor 2, so that the fractions have the lowest common denominator \displaystyle 2\cdot 2\cdot 3\cdot 7,
\displaystyle \begin{align}
\frac{1}{12}-\frac{1}{14} &= \frac{1}{2\cdot 2\cdot 3}-\frac{1}{2\cdot 7}\\[5pt] &= \frac{1}{2\cdot 2\cdot 3}\cdot \frac{7}{7}-\frac{1}{2\cdot 7}\cdot \frac{2\cdot 3}{2\cdot 3}\\[5pt] &= \frac{7}{2\cdot 2\cdot 3\cdot 7} - \frac{2\cdot 3}{2\cdot 2\cdot 3\cdot 7}\\[5pt] &= \frac{7}{84} - \frac{6}{84}\,\textrm{.} \end{align} |
The lowest common denominator is 84.