Lösung 2.1:7c

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We multiply the top and bottom of the first term by <math>a+1</math>, so that both terms then have the same denominator,
We multiply the top and bottom of the first term by <math>a+1</math>, so that both terms then have the same denominator,
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{{Displayed math||<math>\frac{ax}{a+1}\cdot\frac{a+1}{a+1} - \frac{ax^{2}}{(a+1)^{2}} = \frac{ax(a+1)-ax^{2}}{(a+1)^{2}}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{ax}{a+1}\cdot\frac{a+1}{a+1} - \frac{ax^{2}}{(a+1)^{2}} = \frac{ax(a+1)-ax^{2}}{(a+1)^{2}}\,\textrm{.}</math>}}
Because both terms in the numerator contain the factor <math>ax</math>, we take out that factor, obtaining
Because both terms in the numerator contain the factor <math>ax</math>, we take out that factor, obtaining
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{{Displayed math||<math>\frac{ax(a+1-x)}{(a+1)^{2}}</math>}}
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{{Abgesetzte Formel||<math>\frac{ax(a+1-x)}{(a+1)^{2}}</math>}}
and see that the answer cannot be simplified any further.
and see that the answer cannot be simplified any further.
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Note: It is only factors in the numerator and denominator that can cancel each other out, not individual terms. Hence, the following "cancellation" is wrong
Note: It is only factors in the numerator and denominator that can cancel each other out, not individual terms. Hence, the following "cancellation" is wrong
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{{Displayed math||<math>\frac{ax(\rlap{/\,/\,/\,/}a+1)-ax^2}{(a+1)^{2\llap{/}}} = \frac{ax-ax^{2}}{a+1}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{ax(\rlap{/\,/\,/\,/}a+1)-ax^2}{(a+1)^{2\llap{/}}} = \frac{ax-ax^{2}}{a+1}\,\textrm{.}</math>}}

Version vom 08:25, 22. Okt. 2008

We multiply the top and bottom of the first term by \displaystyle a+1, so that both terms then have the same denominator,

\displaystyle \frac{ax}{a+1}\cdot\frac{a+1}{a+1} - \frac{ax^{2}}{(a+1)^{2}} = \frac{ax(a+1)-ax^{2}}{(a+1)^{2}}\,\textrm{.}

Because both terms in the numerator contain the factor \displaystyle ax, we take out that factor, obtaining

\displaystyle \frac{ax(a+1-x)}{(a+1)^{2}}

and see that the answer cannot be simplified any further.

Note: It is only factors in the numerator and denominator that can cancel each other out, not individual terms. Hence, the following "cancellation" is wrong

\displaystyle \frac{ax(\rlap{/\,/\,/\,/}a+1)-ax^2}{(a+1)^{2\llap{/}}} = \frac{ax-ax^{2}}{a+1}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:7c“