Lösung 3.4:1a

Aus Online Mathematik Brückenkurs 2

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The numerator can be factorized using the formula for the difference of two squares to give <math>x^2-1=(x+1)(x-1)</math> and then we see that the numerator and denominator have a common factor which we can eliminate
The numerator can be factorized using the formula for the difference of two squares to give <math>x^2-1=(x+1)(x-1)</math> and then we see that the numerator and denominator have a common factor which we can eliminate
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{{Displayed math||<math>\frac{x^{2}-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{x^{2}-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1\,\textrm{.}</math>}}

Version vom 13:14, 10. 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{.}