Lösung 2.1:7b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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The denominators <math>x-1</math> and <math>x^{2}</math> do not have a common denominator, so the lowest common denominator is <math>x^{2}(x-1)</math>. We treat all three terms so that they have a common denominator and then start simplifying | The denominators <math>x-1</math> and <math>x^{2}</math> do not have a common denominator, so the lowest common denominator is <math>x^{2}(x-1)</math>. We treat all three terms so that they have a common denominator and then start simplifying | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
x + \frac{1}{x-1} + \frac{1}{x^{2}} | x + \frac{1}{x-1} + \frac{1}{x^{2}} | ||
&= x\cdot\frac{x^{2}(x-1)}{x^{2}(x-1)} + \frac{1}{x-1}\cdot\frac{x^{2}}{x^{2}} + \frac{1}{x^{2}}\cdot\frac{x-1}{x-1}\\[5pt] | &= x\cdot\frac{x^{2}(x-1)}{x^{2}(x-1)} + \frac{1}{x-1}\cdot\frac{x^{2}}{x^{2}} + \frac{1}{x^{2}}\cdot\frac{x-1}{x-1}\\[5pt] | ||
Version vom 08:25, 22. Okt. 2008
The denominators \displaystyle x-1 and \displaystyle x^{2} do not have a common denominator, so the lowest common denominator is \displaystyle x^{2}(x-1). We treat all three terms so that they have a common denominator and then start simplifying
| \displaystyle \begin{align}
x + \frac{1}{x-1} + \frac{1}{x^{2}} &= x\cdot\frac{x^{2}(x-1)}{x^{2}(x-1)} + \frac{1}{x-1}\cdot\frac{x^{2}}{x^{2}} + \frac{1}{x^{2}}\cdot\frac{x-1}{x-1}\\[5pt] &= \frac{x^{3}(x-1)+x^{2}+(x-1)}{x^{2}(x-1)}\\[5pt] &= \frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}(x-1)}\,\textrm{.} \end{align} |
