Lösung 2.1:7b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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| - | The denominators | + | 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 |
| - | <math>x-1</math> | + | |
| - | and | + | |
| - | <math>x^{2}</math> | + | |
| - | do not have a common denominator, so the lowest common denominator is | + | |
| - | <math>x^{2} | + | |
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\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} | + | &= \frac{x^{3}(x-1)+x^{2}+(x-1)}{x^{2}(x-1)}\\[5pt] |
| - | \end{align}</math> | + | &= \frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}(x-1)}\,\textrm{.} |
| + | \end{align}</math>}} | ||
Version vom 12:00, 23. Sep. 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
