Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 09:24, 31 March 2008 (edit) Lina (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_1d.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (07:48, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	After <math> x^3y^2 </math> are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator,
-	<center> [[Bild:2_1_1d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	x^3y^2\Big( \frac{1}{y} - \frac{1}{xy} +1 \Big) &= x^3y^2 \cdot\frac{1}{y} -x^3y^2 \cdot \frac{1}{xy} +x^3y^2\cdot 1 \\
		+	&=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\
		+	&=x^3y - x^2y +x^3y^2\,,
		+	\end{align}</math>}}
		+
		+	where we have used
		+
		+	{{Displayed math||<math>\begin{align}
		+	\frac{x^3y^2}{y} &= \frac{x^3\cdot y\cdot{}\rlap{/}y}{\rlap{/}y}= x^3y\,,\\[5pt]
		+	\frac{x^3y^2}{xy} &= \frac{\rlap{/}x\cdot x\cdot x \cdot y \cdot {}\rlap{/}y}{\rlap{/}x\cdot {}\rlap{/}y} = x\cdot x\cdot y = x^2y\,\textrm{.}\end{align}</math>}}

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Current revision

After x3y2 are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator,

x3y2y1−1xy+1=x3y2y1−x3y21xy+x3y21=yx3y2−xyx3y2+x3y2=x3y−x2y+x3y2

where we have used

yx3y2xyx3y2=yx3yy=x3y=xyxxxyy=xxy=x2y.