From Förberedande kurs i matematik 1

Revision as of 11:21, 31 March 2008 (edit) Lina (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_7c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (12:10, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We multiply the top and bottom of the first term by <math>a+1</math>, so that both terms then have the same denominator,
-	<center> [[Bild:2_1_7c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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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		+	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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		+	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
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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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Current revision

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{.}