Solution 2.1:5a - Revision history

Tek at 10:35, 23 September 2008

Tek — Tue, 23 Sep 2008 10:35:58 GMT

←Older revision		Revision as of 10:35, 23 September 2008
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-	In the same way that we calculated fractions, we can subtract the terms' numerators if we first	+	In the same way that we calculated fractions, we can subtract the terms' numerators if we first expand the fractions so that they have the same denominator. Because the denominators are <math>x-x^{2}=x(1-x)</math> and <math>x</math>, the lowest common denominator is <math>x(1-x)</math>,
-	expand the fractions so that they have the same denominator. Because the denominators are	+
-	<math>x-x^{2}=x~~\left~~( 1-x ~~\right~~)</math>	+
-	and	+
-	<math>x</math>, the lowest common denominator is	+
-	<math>x~~\left~~( 1-x ~~\right~~)</math>:	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}}
		+	&= \frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}}\cdot \frac{1-x}{1-x\vphantom{x^2}}\\[5pt]
		+	&= \frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}}\\[5pt]
		+	&= \frac{1-(1-x)}{x-x^{2}}\\[5pt]
		+	&= \frac{1-1+x}{x-x^{2}}\\[5pt]
		+	&= \frac{x}{x-x^{2}}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	This fraction can be simplified by eliminating the factor ''x'' from the numerator and denominator
-	~~& \frac{1}{~~x~~-x^{2}}-\frac{1}{x}=\frac{1}{x-x^{2}}-\frac{1}{x}\centerdot \frac{1-x}{1-x}=\frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}} \\~~	+
-	~~& \\~~	+
-	~~& =\frac{1-\left( 1-x \right)}{x-x^{2}}=\frac{1-1+x}{x-x^{2}}=\frac{x}{x-x^{2}} \\~~	+
-	~~& \\~~	+
-	~~\end{align}</math>~~	+

-		+	{{Displayed math\|\|<math>\frac{x}{x-x^{2}} = \frac{x}{x(1-x)} = \frac{1}{1-x}\,\textrm{.}</math>}}
-	~~This fraction can be simplified by eliminating the factor~~	+
-	<math~~>x</math>~~	+
-	~~from the numerator and denominator.~~	+
-		+
-		+
-		+
-	<math>~~\begin{align}~~	+
-	& \frac{x}{x-x^{2}}=\frac{x}{x~~\left~~( 1-x ~~\right~~)}=\frac{1}{1-x}. \\	+
-	~~& \\~~	+
-	~~\end~~{~~align~~}</math>	+

Ian at 14:46, 15 September 2008

Ian — Mon, 15 Sep 2008 14:46:04 GMT

←Older revision		Revision as of 14:46, 15 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	In the same way that we calculated fractions, we can subtract the terms' numerators if we first
-	<~~center~~> ~~[[Image:2_1_5a.gif]]~~ </~~center~~>	+	expand the fractions so that they have the same denominator. Because the denominators are
-	{{~~NAVCONTENT_STOP~~}}	+	<math>x-x^{2}=x\left( 1-x \right)</math>
		+	and
		+	<math>x</math>, the lowest common denominator is
		+	<math>x\left( 1-x \right)</math>:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{x-x^{2}}-\frac{1}{x}=\frac{1}{x-x^{2}}-\frac{1}{x}\centerdot \frac{1-x}{1-x}=\frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}} \\
		+	& \\
		+	& =\frac{1-\left( 1-x \right)}{x-x^{2}}=\frac{1-1+x}{x-x^{2}}=\frac{x}{x-x^{2}} \\
		+	& \\
		+	\end{align}</math>
		+
		+
		+	This fraction can be simplified by eliminating the factor
		+	<math>x</math>
		+	from the numerator and denominator.
		+
		+
		+
		+	<math>\begin{align}
		+	& \frac{x}{x-x^{2}}=\frac{x}{x\left( 1-x \right)}=\frac{1}{1-x}. \\
		+	& \\
		+	\end{align}</math>

Tekbot: Lösning 2.1:5a moved to Solution 2.1:5a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:37:43 GMT

Lösning 2.1:5a moved to Solution 2.1:5a: Robot: moved page

←Older revision

Revision as of 08:37, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:35:59 GMT

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

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Lina — Mon, 31 Mar 2008 10:06:11 GMT

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