Lösung 3.4:1a

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The numerator can be factorized using the conjugate rule to give
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<center> [[Image:3_4_1a.gif]] </center>
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<math>x^{2}-1=\left( x+1 \right)\left( x-1 \right)</math>
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and then we see that the numerator and denominator have a common factor which we can eliminate
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<math>\frac{x^{2}-1}{x-1}=\frac{\left( x+1 \right)\left( x-1 \right)}{x-1}=x+1</math>

Version vom 12:30, 26. Okt. 2008

The numerator can be factorized using the conjugate rule to give \displaystyle x^{2}-1=\left( x+1 \right)\left( x-1 \right) 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{\left( x+1 \right)\left( x-1 \right)}{x-1}=x+1

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1a“