Lösung 2.1:7b

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.1:7b moved to Solution 2.1:7b: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
The denominators
-
<center> [[Image:2_1_7b.gif]] </center>
+
<math>x-1</math>
-
{{NAVCONTENT_STOP}}
+
and
 +
<math>x^{2}</math>
 +
do not have a common denominator, so the lowest common denominator is
 +
<math>x^{2}\left( x-1 \right)</math>. We treat all three terms so that they have a common denominator and then start simplifying:
 +
 
 +
 
 +
<math>\begin{align}
 +
& x+\frac{1}{x-1}+\frac{1}{x^{2}}=x\centerdot \frac{x^{2}\left( x-1 \right)}{x^{2}\left( x-1 \right)}+\frac{1}{x-1}\centerdot \frac{x^{2}}{x^{2}}+\frac{1}{x^{2}}\centerdot \frac{x-1}{x-1} \\
 +
& =\frac{x^{3}\left( x-1 \right)+x^{2}+\left( x-1 \right)}{x^{2}\left( x-1 \right)}=\frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}\left( x-1 \right)} \\
 +
\end{align}</math>

Version vom 12:46, 16. Sep. 2008

The denominators \displaystyle x-1 and \displaystyle x^{2} do not have a common denominator, so the lowest common denominator is \displaystyle x^{2}\left( x-1 \right). We treat all three terms so that they have a common denominator and then start simplifying:


\displaystyle \begin{align} & x+\frac{1}{x-1}+\frac{1}{x^{2}}=x\centerdot \frac{x^{2}\left( x-1 \right)}{x^{2}\left( x-1 \right)}+\frac{1}{x-1}\centerdot \frac{x^{2}}{x^{2}}+\frac{1}{x^{2}}\centerdot \frac{x-1}{x-1} \\ & =\frac{x^{3}\left( x-1 \right)+x^{2}+\left( x-1 \right)}{x^{2}\left( x-1 \right)}=\frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}\left( x-1 \right)} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:7b“