Solution 3.1:7b - Revision history

Tek at 14:15, 30 September 2008

Tek — Tue, 30 Sep 2008 14:15:28 GMT

←Older revision		Revision as of 14:15, 30 September 2008
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	We multiply the top and bottom of the fraction by the conjugate of the denominator,		We multiply the top and bottom of the fraction by the conjugate of the denominator,
-	<math>\sqrt{7}+\sqrt{5}</math>	+	<math>\sqrt{7}+\sqrt{5}</math>, and see what it leads to,
-	, and see what it leads to:	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}
-	& \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\~~centerdot~~ \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{~~\left~~( 5\sqrt{7}-7\sqrt{5} ~~\right~~)~~\centerdot \left~~( \sqrt{7}+\sqrt{5} ~~\right~~)}{~~\left~~( \sqrt{7} ~~\right~~)^{2}-~~\left~~( \sqrt{5} ~~\right~~)^{2}} \\	+	&= \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\[10pt]
-	& =\frac{5\sqrt{7}\~~centerdot~~ \sqrt{7}+5\sqrt{5}\~~centerdot~~ \sqrt{7}-7\sqrt{5}\~~centerdot~~ \sqrt{7}-7\sqrt{5}\~~centerdot~~ \sqrt{5}}{7-5} \\	+	&= \frac{(5\sqrt{7}-7\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\\[10pt]
-	& =\frac{5~~\left~~( \sqrt{7} ~~\right~~)^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7~~\left~~( \sqrt{5} ~~\right~~)^{2}}{2} \\	+	&= \frac{5\sqrt{7}\cdot\sqrt{7}+5\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{5}}{7-5}\\[10pt]
-	& =\frac{5\~~centerdot~~ 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\~~centerdot~~ 5}{2} \\	+	&= \frac{5(\sqrt{7})^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7(\sqrt{5})^{2}}{2}\\[10pt]
-	& =\frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}=\frac{~~\left~~( 5-7 ~~\right~~)\sqrt{5}\sqrt{7}}{2}=\frac{-2\sqrt{5}\~~sqrt{~~7}}{2} \\	+	&= \frac{5\cdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\cdot 5}{2}\\[10pt]
-	& =-\sqrt{35} \\	+	&= \frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}\\[10pt]
-	\end{align}</math>	+	&= \frac{(5-7)\sqrt{5}\sqrt{7}}{2}\\[10pt]
		+	&= \frac{-2\sqrt{5\cdot 7}}{2}\\[10pt]
		+	&= -\sqrt{35}\,\textrm{.}
		+	\end{align}</math>}}

Ian at 10:15, 23 September 2008

Ian — Tue, 23 Sep 2008 10:15:00 GMT

←Older revision		Revision as of 10:15, 23 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We multiply the top and bottom of the fraction by the conjugate of the denominator,
-	<~~center~~> ~~[[Image~~:~~3_1_7b.gif]]~~ <~~/center~~>	+	<math>\sqrt{7}+\sqrt{5}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, and see what it leads to:
		+
		+
		+	<math>\begin{align}
		+	& \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\centerdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left( 5\sqrt{7}-7\sqrt{5} \right)\centerdot \left( \sqrt{7}+\sqrt{5} \right)}{\left( \sqrt{7} \right)^{2}-\left( \sqrt{5} \right)^{2}} \\
		+	& =\frac{5\sqrt{7}\centerdot \sqrt{7}+5\sqrt{5}\centerdot \sqrt{7}-7\sqrt{5}\centerdot \sqrt{7}-7\sqrt{5}\centerdot \sqrt{5}}{7-5} \\
		+	& =\frac{5\left( \sqrt{7} \right)^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\left( \sqrt{5} \right)^{2}}{2} \\
		+	& =\frac{5\centerdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\centerdot 5}{2} \\
		+	& =\frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}=\frac{\left( 5-7 \right)\sqrt{5}\sqrt{7}}{2}=\frac{-2\sqrt{5}\sqrt{7}}{2} \\
		+	& =-\sqrt{35} \\
		+	\end{align}</math>

Tekbot: Lösning 3.1:7b moved to Solution 3.1:7b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:15:23 GMT

Lösning 3.1:7b moved to Solution 3.1:7b: Robot: moved page

←Older revision

Revision as of 09:15, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:54:49 GMT

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

←Older revision		Revision as of 06:54, 21 August 2008
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-	<center> [[~~Bild~~:3_1_7b.gif]] </center>	+	<center> [[Image:3_1_7b.gif]] </center>
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_7b.gif {{NAVCONTENT_STOP}}

Oskar — Tue, 01 Apr 2008 14:13:22 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_7b.gif </center> {{NAVCONTENT_STOP}}

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