Solution 2.1:7b - Revision history

Tek at 12:00, 23 September 2008

2008-09-23T12:00:31Z

←Older revision		Revision as of 12:00, 23 September 2008
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-	The denominators	+	The denominators <math>x-1</math> and <math>x^{2}</math> do not have a common denominator, so the lowest common denominator is <math>x^{2}(x-1)</math>. We treat all three terms so that they have a common denominator and then start simplifying
-	<math>x-1</math>	+
-	and	+
-	<math>x^{2}</math>	+
-	do not have a common denominator, so the lowest common denominator is	+
-	<math>x^{2}~~\left~~( x-1 ~~\right~~)</math>. We treat all three terms so that they have a common denominator and then start simplifying:	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	x + \frac{1}{x-1} + \frac{1}{x^{2}}
-	& x+\frac{1}{x-1}+\frac{1}{x^{2}}=x\~~centerdot~~ \frac{x^{2}~~\left~~( x-1 ~~\right~~)}{x^{2}~~\left~~( x-1 ~~\right~~)}+\frac{1}{x-1}\~~centerdot~~ \frac{x^{2}}{x^{2}}+\frac{1}{x^{2}}\~~centerdot~~ \frac{x-1}{x-1} \\	+	&= x\cdot\frac{x^{2}(x-1)}{x^{2}(x-1)} + \frac{1}{x-1}\cdot\frac{x^{2}}{x^{2}} + \frac{1}{x^{2}}\cdot\frac{x-1}{x-1}\\[5pt]
-	& =\frac{x^{3}~~\left~~( x-1 ~~\right~~)+x^{2}+~~\left~~( x-1 ~~\right~~)}{x^{2}~~\left~~( x-1 ~~\right~~)}=\frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}~~\left~~( x-1 ~~\right~~)} \\	+	&= \frac{x^{3}(x-1)+x^{2}+(x-1)}{x^{2}(x-1)}\\[5pt]
-	\end{align}</math>	+	&= \frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}(x-1)}\,\textrm{.}
		+	\end{align}</math>}}

2008-09-16T12:46:27Z

←Older revision		Revision as of 12:46, 16 September 2008
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-	{~~{NAVCONTENT_START}~~}	+	The denominators
-	<~~center~~> ~~[[Image:2_1_7b.gif]]~~ </~~center~~>	+	<math>x-1</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>x^{2}</math>
		+	do not have a common denominator, so the lowest common denominator is
		+	<math>x^{2}\left( x-1 \right)</math>. We treat all three terms so that they have a common denominator and then start simplifying:
		+
		+
		+	<math>\begin{align}
		+	& x+\frac{1}{x-1}+\frac{1}{x^{2}}=x\centerdot \frac{x^{2}\left( x-1 \right)}{x^{2}\left( x-1 \right)}+\frac{1}{x-1}\centerdot \frac{x^{2}}{x^{2}}+\frac{1}{x^{2}}\centerdot \frac{x-1}{x-1} \\
		+	& =\frac{x^{3}\left( x-1 \right)+x^{2}+\left( x-1 \right)}{x^{2}\left( x-1 \right)}=\frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}\left( x-1 \right)} \\
		+	\end{align}</math>

2008-09-09T08:40:43Z

←Older revision

Revision as of 08:40, 9 September 2008

2008-08-21T06:37:29Z

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2008-03-31T11:19:39Z

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