Lösung 1.2:5a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
| Zeile 1: | Zeile 1: | ||
We begin by calculating the numerator in the main fraction, | We begin by calculating the numerator in the main fraction, | ||
| - | {{ | + | {{Abgesetzte Formel||<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>.}} |
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>, | 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>, | ||
| - | {{ | + | {{Abgesetzte Formel||<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>.}} |
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 | 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 | ||
| - | {{ | + | {{Abgesetzte Formel||<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>.}} |
Version vom 08:15, 22. Okt. 2008
We begin by calculating the numerator in the main fraction,
| \displaystyle \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(}\,}\,. |
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,
| \displaystyle \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}\,. |
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
| \displaystyle \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}\,. |
