Lösung 2.1:7c

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K (Lösning 2.1:7c moved to Solution 2.1:7c: Robot: moved page)
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We multiply the top and bottom of the first term by
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<center> [[Image:2_1_7c.gif]] </center>
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<math>a+1</math>, so that both terms then have the same
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denominator,
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<math>\frac{ax}{a+1}\centerdot \frac{a+1}{a+1}-\frac{ax^{2}}{\left( a+1 \right)^{2}}=\frac{ax\left( a+1 \right)-ax^{2}}{\left( a+1 \right)^{2}}</math>
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Because both terms in the numerator contain the factor
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<math>ax</math>, we take out that factor, obtaining
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<math>\frac{ax\left( a+1-x \right)}{\left( a+1 \right)^{2}}</math>
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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
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individual terms. Hence, the following "cancellation" is wrong:
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<math>\frac{ax\left( a+1-x \right)}{\left( a+1 \right)^{2}}=\frac{ax-ax^{2}}{a+1}</math>

Version vom 13:02, 16. 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,


\displaystyle \frac{ax}{a+1}\centerdot \frac{a+1}{a+1}-\frac{ax^{2}}{\left( a+1 \right)^{2}}=\frac{ax\left( a+1 \right)-ax^{2}}{\left( a+1 \right)^{2}}


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


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

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:7c“