Solution 3.4:1a

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Current revision (11:44, 31 October 2008) (edit) (undo)
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The numerator can be factorized using the conjugate rule to give
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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
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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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{{Displayed math||<math>\frac{x^{2}-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1\,\textrm{.}</math>}}
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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>
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Current revision

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{.}
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