Lösung 2.1:7a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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If we multiply the top and bottom of the first fraction by <math>x+5</math> and the second by <math>x+3</math>, then they will both have the same denominator and we can work out the expression by subtracting the numerators | If we multiply the top and bottom of the first fraction by <math>x+5</math> and the second by <math>x+3</math>, then they will both have the same denominator and we can work out the expression by subtracting the numerators | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{2}{x+3}-\frac{2}{x+5} &= \frac{2}{x+3}\cdot \frac{x+5}{x+5}-\frac{2}{x+5}\cdot \frac{x+3}{x+3}\\[7pt] | \frac{2}{x+3}-\frac{2}{x+5} &= \frac{2}{x+3}\cdot \frac{x+5}{x+5}-\frac{2}{x+5}\cdot \frac{x+3}{x+3}\\[7pt] | ||
&= \frac{2(x+5)-2(x+3)}{(x+3)(x+5)}\\[5pt] | &= \frac{2(x+5)-2(x+3)}{(x+3)(x+5)}\\[5pt] | ||
Version vom 08:25, 22. Okt. 2008
If we multiply the top and bottom of the first fraction by \displaystyle x+5 and the second by \displaystyle x+3, then they will both have the same denominator and we can work out the expression by subtracting the numerators
| \displaystyle \begin{align}
\frac{2}{x+3}-\frac{2}{x+5} &= \frac{2}{x+3}\cdot \frac{x+5}{x+5}-\frac{2}{x+5}\cdot \frac{x+3}{x+3}\\[7pt] &= \frac{2(x+5)-2(x+3)}{(x+3)(x+5)}\\[5pt] &= \frac{2x+10-2x-6}{(x+3)(x+5)}\\[5pt] &= \frac{4}{(x+3)(x+5)}\,\textrm{.} \end{align} |
