Lösung 3.4:1a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
The numerator can be factorized using the conjugate rule to give
+
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
-
<math>x^{2}-1=\left( x+1 \right)\left( x-1 \right)</math>
+
-
and then we see that the numerator and denominator have a common factor which we can eliminate
+
-
 
+
{{Displayed math||<math>\frac{x^{2}-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1\,\textrm{.}</math>}}
-
<math>\frac{x^{2}-1}{x-1}=\frac{\left( x+1 \right)\left( x-1 \right)}{x-1}=x+1</math>
+

Version vom 11:44, 31. Okt. 2008

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