Solution 2.1:1d - Revision history

Tek at 07:48, 23 September 2008

2008-09-23T07:48:58Z

←Older revision		Revision as of 07:48, 23 September 2008
Line 1:		Line 1:
-	~~{{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,
-	After <math> x^3y^2 </math> are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator.	+

-	<math> ~~\qquad~~ \begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	x^3y^2\Big( \frac 1y - \frac 1{xy} +1 \Big) &= x^3y^2 \cdot\frac 1y -x^3y^2 \cdot \frac 1{xy} +x^3y^2\cdot 1 \\	+	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 \\		&=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\
-	&=x^3y - x^2y +x^3y^2	+	&=x^3y - x^2y +x^3y^2\,,
-	\end{align}</math>	+	\end{align}</math>}}

	where we have used		where we have used

-	<math> \~~qquad~~ \frac{x^3y^2}{y}= \frac{x^3\cdot y\cdot y}{y}= x^3y ~~</math>~~,	+	{{Displayed math\|\|<math>\begin{align}
-		+	\frac{x^3y^2}{y} &= \frac{x^3\cdot y\cdot{}\rlap{/}y}{\rlap{/}y}= x^3y\,,\\[5pt]
-	~~<math>~~\~~qquad~~ \frac{x^3y^2}{xy}=\frac{x\cdot x\cdot x \cdot y \cdot y}{x\cdot y} = x\cdot x\cdot y = x^2y </math>	+	\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>}}
-	~~<!-- <center> [[Image:2_1_1d.gif]] </center>-->~~	+
-	~~{{NAVCONTENT_STOP~~}}	+

2008-09-09T08:31:23Z

←Older revision

Revision as of 08:31, 9 September 2008

2008-08-21T06:32:49Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

←Older revision		Revision as of 06:32, 21 August 2008
Line 13:		Line 13:

	<math>\qquad \frac{x^3y^2}{xy}=\frac{x\cdot x\cdot x \cdot y \cdot y}{x\cdot y} = x\cdot x\cdot y = x^2y </math>		<math>\qquad \frac{x^3y^2}{xy}=\frac{x\cdot x\cdot x \cdot y \cdot y}{x\cdot y} = x\cdot x\cdot y = x^2y </math>
-	<!-- <center> [[~~Bild~~:2_1_1d.gif]] </center>-->	+	<!-- <center> [[Image:2_1_1d.gif]] </center>-->
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

2008-08-13T08:57:03Z

←Older revision		Revision as of 08:57, 13 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:2_1_1d.gif]] </center>	+	After <math> x^3y^2 </math> are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator.
		+
		+	<math> \qquad \begin{align}
		+	x^3y^2\Big( \frac 1y - \frac 1{xy} +1 \Big) &= x^3y^2 \cdot\frac 1y -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
		+
		+	<math> \qquad \frac{x^3y^2}{y}= \frac{x^3\cdot y\cdot y}{y}= x^3y </math>,
		+
		+	<math>\qquad \frac{x^3y^2}{xy}=\frac{x\cdot x\cdot x \cdot y \cdot y}{x\cdot y} = x\cdot x\cdot y = x^2y </math>
		+	<!-- <center> [[Bild:2_1_1d.gif]] </center>-->
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

2008-03-31T09:24:36Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_1d.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:2_1_1d.gif]] </center>
{{NAVCONTENT_STOP}}