Lösung 2.1:8a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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In our expression, we multiply the top and bottom of the main fraction by <math>x+1</math> (so as to get rid of <math>x+1</math> from the numerator), | In our expression, we multiply the top and bottom of the main fraction by <math>x+1</math> (so as to get rid of <math>x+1</math> from the numerator), | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\frac{\,\dfrac{x}{x+1}\,}{3+x} = \frac{\,\dfrac{x}{x+1}\,}{3+x}\cdot\frac{x+1}{x+1} = \frac{\,\dfrac{x}{x+1}\cdot (x+1)\,}{\,(3+x)(x+1)\,} = \frac{x}{(3+x)(x+1)}\,\textrm{.}</math>}} |
Version vom 08:25, 22. Okt. 2008
An expression which consists of several fraction signs can be rewritten in terms of one fraction sign by systematically eliminating all partial fractions.
In our expression, we multiply the top and bottom of the main fraction by \displaystyle x+1 (so as to get rid of \displaystyle x+1 from the numerator),
| \displaystyle \frac{\,\dfrac{x}{x+1}\,}{3+x} = \frac{\,\dfrac{x}{x+1}\,}{3+x}\cdot\frac{x+1}{x+1} = \frac{\,\dfrac{x}{x+1}\cdot (x+1)\,}{\,(3+x)(x+1)\,} = \frac{x}{(3+x)(x+1)}\,\textrm{.} |
