Lösung 1.2:5c
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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{{Displayed math||<math>\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\vphantom{\Biggl(}\,}{\,\dfrac{11}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\cdot \dfrac{16}{11}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{\,/}11}{\rlap{\,/}16}\cdot \dfrac{\rlap{\,/}16}{\rlap{\,/}11}\vphantom{\Biggl(}\,} = \dfrac{16}{10\cdot 11}</math>}} | {{Displayed math||<math>\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\vphantom{\Biggl(}\,}{\,\dfrac{11}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\cdot \dfrac{16}{11}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{\,/}11}{\rlap{\,/}16}\cdot \dfrac{\rlap{\,/}16}{\rlap{\,/}11}\vphantom{\Biggl(}\,} = \dfrac{16}{10\cdot 11}</math>}} | ||
| - | and because <math>16=2\cdot 2\cdot 2\cdot 2</math> and <math>10=2\ | + | and because <math>16=2\cdot 2\cdot 2\cdot 2</math> and <math>10=2\cdot 5</math>, the simplified answer is |
{{Displayed math||<math>\frac{16}{10\cdot 11} = \frac{\rlap{/}2\cdot 2\cdot 2\cdot 2}{\rlap{/}2\cdot 5\cdot 11} = \frac{8}{55}\,</math>.}} | {{Displayed math||<math>\frac{16}{10\cdot 11} = \frac{\rlap{/}2\cdot 2\cdot 2\cdot 2}{\rlap{/}2\cdot 5\cdot 11} = \frac{8}{55}\,</math>.}} | ||
Version vom 12:34, 22. Sep. 2008
Method 1
We calculate the numerator and denominator first
Thus, the expression becomes
and because \displaystyle 16=2\cdot 2\cdot 2\cdot 2 and \displaystyle 10=2\cdot 5, the simplified answer is
Method 2
If we look at the individual fractions 3/10, 1/5, 7/8 and 3/16, we see that the denominators can be factorized as
and therefore 2∙2∙2∙2∙5 = 80 is the fractions' lowest common denominator.
If we multiply the top and bottom of the main fraction by 80, then it will be possible to eliminate all denominators at once,
