Lösung 2.1:7c
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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- | We multiply the top and bottom of the first term by | + | We multiply the top and bottom of the first term by <math>a+1</math>, so that both terms then have the same denominator, |
- | <math>a+1</math>, so that both terms then have the same | + | |
- | denominator, | + | |
+ | {{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>}} | ||
- | <math> | + | Because both terms in the numerator contain the factor <math>ax</math>, we take out that factor, obtaining |
- | + | {{Displayed math||<math>\frac{ax(a+1-x)}{(a+1)^{2}}</math>}} | |
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- | <math>\frac{ax | + | |
and see that the answer cannot be simplified any further. | 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 | |
- | individual terms. Hence, the following "cancellation" is wrong | + | |
- | + | ||
- | <math>\frac{ax | + | {{Displayed math||<math>\frac{ax(\rlap{/\,/\,/\,/}a+1)-ax^2}{(a+1)^{2\llap{/}}} = \frac{ax-ax^{2}}{a+1}\,\textrm{.}</math>}} |
Version vom 12:10, 23. Sep. 2008
We multiply the top and bottom of the first term by \displaystyle a+1, so that both terms then have the same denominator,
Because both terms in the numerator contain the factor \displaystyle ax, we take out that factor, obtaining
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