Lösung 2.1:6c

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K (Lösning 2.1:6c moved to Solution 2.1:6c: Robot: moved page)
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Because the numerators are
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<center> [[Image:2_1_6c.gif]] </center>
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<math>a^{2}-ab=a\left( a-b \right)</math>
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and
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<math>a-b</math>, both terms will have a common denominator
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<math>a\left( a-b \right)</math>
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if the top and bottom of the second term are multiplied by
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<math>a</math>:
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<math>\begin{align}
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& \frac{2a+b}{a^{2}-b}-\frac{2}{a-b}=\frac{2a+b}{a\left( a-b \right)}-\frac{2}{a-b}\centerdot \frac{a}{a} \\
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& =\frac{2a+b-2a}{a\left( a-b \right)}=\frac{b}{a\left( a-b \right)} \\
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\end{align}</math>

Version vom 11:03, 16. Sep. 2008

Because the numerators are \displaystyle a^{2}-ab=a\left( a-b \right) and \displaystyle a-b, both terms will have a common denominator \displaystyle a\left( a-b \right) if the top and bottom of the second term are multiplied by \displaystyle a:


\displaystyle \begin{align} & \frac{2a+b}{a^{2}-b}-\frac{2}{a-b}=\frac{2a+b}{a\left( a-b \right)}-\frac{2}{a-b}\centerdot \frac{a}{a} \\ & =\frac{2a+b-2a}{a\left( a-b \right)}=\frac{b}{a\left( a-b \right)} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:6c“