Lösung 3.4:1a
Aus Online Mathematik Brückenkurs 2
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Version vom 10:50, 11. Mär. 2009
The numerator can be factorized using the formula for the difference of two squares to give \displaystyle x^2-1=(x+1)(x-1) and then we see that the numerator and denominator have a common factor which we can eliminate
| \displaystyle \frac{x^{2}-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1\,\textrm{.} |
