Lösung 2.1:6c
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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Because the denominators are <math>a^{2}-ab = a(a-b)</math> and <math>a-b</math>, both terms will have a common denominator <math>a(a-b)</math> if the top and bottom of the second term are multiplied by <math>a</math>, | Because the denominators are <math>a^{2}-ab = a(a-b)</math> and <math>a-b</math>, both terms will have a common denominator <math>a(a-b)</math> if the top and bottom of the second term are multiplied by <math>a</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{2a+b}{a^{2}-b}-\frac{2}{a-b} &= \frac{2a+b}{a(a-b)}-\frac{2}{a-b}\cdot\frac{a}{a}\\[5pt] | \frac{2a+b}{a^{2}-b}-\frac{2}{a-b} &= \frac{2a+b}{a(a-b)}-\frac{2}{a-b}\cdot\frac{a}{a}\\[5pt] | ||
&= \frac{2a+b-2a}{a(a-b)}\\[5pt] | &= \frac{2a+b-2a}{a(a-b)}\\[5pt] | ||
&= \frac{b}{a(a-b)}\,\textrm{.} | &= \frac{b}{a(a-b)}\,\textrm{.} | ||
\end{align}</math>}} | \end{align}</math>}} | ||
Version vom 08:24, 22. Okt. 2008
Because the denominators are \displaystyle a^{2}-ab = a(a-b) and \displaystyle a-b, both terms will have a common denominator \displaystyle a(a-b) 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(a-b)}-\frac{2}{a-b}\cdot\frac{a}{a}\\[5pt] &= \frac{2a+b-2a}{a(a-b)}\\[5pt] &= \frac{b}{a(a-b)}\,\textrm{.} \end{align} |
