Solution 3.1:3d - Revision history

Tek at 10:40, 30 September 2008

2008-09-30T10:40:34Z

←Older revision		Revision as of 10:40, 30 September 2008
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-	We can multiply	+	We can multiply <math>\sqrt{\tfrac{2}{3}}</math> into the bracket and then write the root expressions together under a common root sign using the rule <math>\sqrt{a\vphantom{b}}\cdot \sqrt{b} = \sqrt{ab}</math>,
-	<math>\sqrt{\~~frac~~{2}{3}}</math>	+
-	into the bracket and then write the root expressions together under a common root sign using the rule	+
-	<math>\sqrt{a}\~~centerdot~~ \sqrt{b}=\sqrt{ab}</math>	+

		+	{{Displayed math\|\|<math>\sqrt{\frac{2}{3}}\bigl(\sqrt{6}-\sqrt{3}\bigr) = \sqrt{\frac{2}{3}}\cdot\sqrt{6} - \sqrt{\frac{2}{3}}\cdot\sqrt{3} = \sqrt{\frac{2\cdot 6}{3}} - \sqrt{\frac{2\cdot 3}{3}}\,\textrm{.}</math>}}

		+	Because <math>(2\cdot 6)/3 = 2\cdot 2 = 2^2</math> and <math>(2\cdot 3)/3 = 2</math>, we obtain

-	<math>\sqrt{\frac{2}{3}}\~~left~~( \sqrt{6}-\sqrt{3} \~~right~~)=\sqrt~~{\frac~~{2~~}{3}}\centerdot \sqrt{6~~}-\sqrt~~{\frac~~{2~~}{3}}\centerdot \sqrt{3~~}=~~\sqrt{\frac{~~2~~\centerdot 6}{3}}~~-\sqrt~~{\frac~~{2~~\centerdot 3~~}~~{3}}.</math>~~	+	{{Displayed math\|\|<math>\sqrt{\frac{2}{3}}\bigl(\sqrt{6} - \sqrt{3}\bigr) = \sqrt{2^2}-\sqrt{2} = 2-\sqrt{2}\,\textrm{.}</math>}}
-		+
-	~~Because~~	+
-	~~<math>~~\~~frac{2~~\~~centerdot 6}~~{~~3}=2\centerdot 2=2^{2~~}</math>	+
-	~~and~~	+
-	~~<math>\frac{2\centerdot 3~~}{3}~~=2</math>, we obtain~~	+
-		+
-		+
-	~~<math>\sqrt{\frac{2}{3}}\left( \sqrt{6}-\sqrt{3} \right)=\sqrt{2^{2}}-\sqrt{2}=2-\sqrt{2}</math>~~	+

2008-09-22T13:07:40Z

←Older revision		Revision as of 13:07, 22 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We can multiply
-	<~~center~~> ~~[[Image:3_1_3d~~.~~gif]]~~ </~~center~~>	+	<math>\sqrt{\frac{2}{3}}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	into the bracket and then write the root expressions together under a common root sign using the rule
		+	<math>\sqrt{a}\centerdot \sqrt{b}=\sqrt{ab}</math>
		+
		+
		+
		+	<math>\sqrt{\frac{2}{3}}\left( \sqrt{6}-\sqrt{3} \right)=\sqrt{\frac{2}{3}}\centerdot \sqrt{6}-\sqrt{\frac{2}{3}}\centerdot \sqrt{3}=\sqrt{\frac{2\centerdot 6}{3}}-\sqrt{\frac{2\centerdot 3}{3}}.</math>
		+
		+	Because
		+	<math>\frac{2\centerdot 6}{3}=2\centerdot 2=2^{2}</math>
		+	and
		+	<math>\frac{2\centerdot 3}{3}=2</math>, we obtain
		+
		+
		+	<math>\sqrt{\frac{2}{3}}\left( \sqrt{6}-\sqrt{3} \right)=\sqrt{2^{2}}-\sqrt{2}=2-\sqrt{2}</math>

2008-09-09T09:10:43Z

←Older revision

Revision as of 09:10, 9 September 2008

2008-08-21T06:52:29Z

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2008-04-01T14:02:27Z

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