Lösung 2.1:5a
Aus Online Mathematik Brückenkurs 1
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Version vom 13:48, 22. Okt. 2008
In the same way that we calculated fractions, we can subtract the terms' numerators if we first expand the fractions so that they have the same denominator. Because the denominators are \displaystyle x-x^{2}=x(1-x) and \displaystyle x, the lowest common denominator is \displaystyle x(1-x),
| \displaystyle \begin{align}
\frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}} &= \frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}}\cdot \frac{1-x}{1-x\vphantom{x^2}}\\[5pt] &= \frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}}\\[5pt] &= \frac{1-(1-x)}{x-x^{2}}\\[5pt] &= \frac{1-1+x}{x-x^{2}}\\[5pt] &= \frac{x}{x-x^{2}}\,\textrm{.} \end{align} |
This fraction can be simplified by eliminating the factor x from the numerator and denominator
| \displaystyle \frac{x}{x-x^{2}} = \frac{x}{x(1-x)} = \frac{1}{1-x}\,\textrm{.} |
