Solution 2.1:5b - Revision history

Tek at 10:51, 23 September 2008

2008-09-23T10:51:55Z

←Older revision		Revision as of 10:51, 23 September 2008
Line 1:		Line 1:
	We can factorize the denominators as		We can factorize the denominators as

		+	{{Displayed math\|\|<math>\begin{align}
		+	y^{2}-2y &= y(y-2)\\
		+	y^{2}-4 &= (y-2)(y+2)\quad\text{[difference of two squares]}
		+	\end{align}</math>}}

-	<math>y~~^{2}~~-~~2y=~~y~~\left~~( y-2 ~~\right~~)</math>	+	and then we see that the terms' lowest common denominator is <math>y(y-2)(y+2)</math> because it is the product that contains the smallest number of factors which contain both <math>y(y-2)</math> and <math>(y-2)(y+2)</math>.

		+	Now, we rewrite the fractions so that they have same denominators and start simplifying

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

		+	The numerator can be rewritten as <math>-y+2=-(y-2)</math> and we can eliminate the common factor <math>y-2</math>,

-	~~<math>y^~~{~~2}-4=\left( y-2 \right)\left( y+2 \right)</~~math>	+	{{Displayed math\|\|<math>\frac{-y+2}{y(y-2)(y+2)} = \frac{-(y-2)}{y(y-2)(y+2)} = \frac{-1}{y(y+2)} = -\frac{1}{y(y+2)}\,\textrm{.}</math>}}
-	~~[by the conjugate rule]~~	+
-		+
-	~~and then we see that the terms' lowest common denominator is~~	+
-	<math>~~y\left( y-2 \right)\left( y+2 \right)</math>~~	+
-	~~because it is the product that contains the smallest number of factors which contain both~~	+
-	~~<math>y\left( y-2 \right)</math>~~	+
-	~~and~~	+
-	~~<math>\left( y-2 \right)\left( y+2 \right)</math>~~	+
-	.	+
-		+
-	~~Now, we rewrite the fractions so that they have same denominators and start simplifying:~~	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& y^{2}-4=\left( y-2 \right)\left( y+2 \right) \\~~	+
-	~~& y^{2}-2y=y\left( y-2 \right) \\~~	+
-	& \frac{~~1}{y^{2}~~-~~2y}-\frac{2}{y^{2}-4}=\frac{1}{y\left( y-2 \right)}\centerdot \frac{~~y+2}{y~~+2}-\frac{2}{\left~~( y-2 ~~\right~~)~~\left~~( y+2 ~~\right~~)}~~\centerdot \frac{y}{y} \\~~	+
-	~~& \\~~	+
-	& =\frac{~~y+2}{y\left( y~~-~~2 \right)\left( y+2 \right)}-\frac{2y}{\left~~( y-2 ~~\right~~)~~\left( y+2 \right)y~~} \\	+
-	~~& \\~~	+
-	~~& =\frac~~{y~~+2-2y}{y\left~~( y-2 ~~\right~~)~~\left~~( y+2 ~~\right~~)}=\frac{-~~y+2~~}{y~~\left( y-2 \right)\left~~( y+2 ~~\right~~)} \\	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~The numerator can be rewritten as~~	+
-	~~<math>-y+2~~=-~~\left( y-2 \right)</math>~~	+
-	~~and we can eliminate the common factor~~	+
-	~~<math>y-2</math>~~	+
-	.	+
-		+
-		+
-	~~<math>~~\frac{~~-y+2~~}{y~~\left( y-2 \right)\left~~( y+2 ~~\right~~)}=\~~frac{-~~\~~left( y-2 \right)}~~{~~y\left( y-2 \right)\left( y+2 \right)}=\frac{-1}{y\left( y+2 \right)}=-\frac{1}{y\left( y+2 \right)~~}</math>	+

Ian at 09:15, 16 September 2008

2008-09-16T09:15:00Z

←Older revision		Revision as of 09:15, 16 September 2008
Line 1:		Line 1:
-	{~~{NAVCONTENT_START}~~}	+	We can factorize the denominators as
-	<~~center~~> [~~[Image:2_1_5b.gif]~~] </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>y^{2}-2y=y\left( y-2 \right)</math>
		+
		+
		+
		+
		+	<math>y^{2}-4=\left( y-2 \right)\left( y+2 \right)</math>
		+	[by the conjugate rule]
		+
		+	and then we see that the terms' lowest common denominator is
		+	<math>y\left( y-2 \right)\left( y+2 \right)</math>
		+	because it is the product that contains the smallest number of factors which contain both
		+	<math>y\left( y-2 \right)</math>
		+	and
		+	<math>\left( y-2 \right)\left( y+2 \right)</math>
		+	.
		+
		+	Now, we rewrite the fractions so that they have same denominators and start simplifying:
		+
		+
		+	<math>\begin{align}
		+	& y^{2}-4=\left( y-2 \right)\left( y+2 \right) \\
		+	& y^{2}-2y=y\left( y-2 \right) \\
		+	& \frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}=\frac{1}{y\left( y-2 \right)}\centerdot \frac{y+2}{y+2}-\frac{2}{\left( y-2 \right)\left( y+2 \right)}\centerdot \frac{y}{y} \\
		+	& \\
		+	& =\frac{y+2}{y\left( y-2 \right)\left( y+2 \right)}-\frac{2y}{\left( y-2 \right)\left( y+2 \right)y} \\
		+	& \\
		+	& =\frac{y+2-2y}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)} \\
		+	\end{align}</math>
		+
		+
		+	The numerator can be rewritten as
		+	<math>-y+2=-\left( y-2 \right)</math>
		+	and we can eliminate the common factor
		+	<math>y-2</math>
		+	.
		+
		+
		+	<math>\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-\left( y-2 \right)}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-1}{y\left( y+2 \right)}=-\frac{1}{y\left( y+2 \right)}</math>

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

2008-09-09T08:38:03Z

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

←Older revision

Revision as of 08:38, 9 September 2008

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

2008-08-21T06:36:09Z

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

←Older revision		Revision as of 06:36, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_1_5b.gif]] </center>	+	<center> [[Image:2_1_5b.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:2_1_5b.gif {{NAVCONTENT_STOP}}

2008-03-31T11:05:44Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_5b.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:2_1_5b.gif]] </center>
{{NAVCONTENT_STOP}}