From Förberedande kurs i matematik 1

Revision as of 08:41, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.1:8a moved to Solution 2.1:8a: Robot: moved page) ← Previous diff		Current revision (12:29, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	An expression which consists of several fraction signs can be rewritten in terms of one fraction sign by systematically eliminating all partial fractions.
-	<center> [[Image:2_1_8a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	In our expression, we multiply the top and bottom of the main fraction by <math>x+1</math> (so as to get rid of <math>x+1</math> from the numerator),
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		+	{{Displayed math||<math>\frac{\,\dfrac{x}{x+1}\,}{3+x} = \frac{\,\dfrac{x}{x+1}\,}{3+x}\cdot\frac{x+1}{x+1} = \frac{\,\dfrac{x}{x+1}\cdot (x+1)\,}{\,(3+x)(x+1)\,} = \frac{x}{(3+x)(x+1)}\,\textrm{.}</math>}}

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Current revision

An expression which consists of several fraction signs can be rewritten in terms of one fraction sign by systematically eliminating all partial fractions.

In our expression, we multiply the top and bottom of the main fraction by \displaystyle x+1 (so as to get rid of \displaystyle x+1 from the numerator),

\displaystyle \frac{\,\dfrac{x}{x+1}\,}{3+x} = \frac{\,\dfrac{x}{x+1}\,}{3+x}\cdot\frac{x+1}{x+1} = \frac{\,\dfrac{x}{x+1}\cdot (x+1)\,}{\,(3+x)(x+1)\,} = \frac{x}{(3+x)(x+1)}\,\textrm{.}