Solution 3.1:6c - Revision history

Tek at 12:49, 30 September 2008

Tek — Tue, 30 Sep 2008 12:49:55 GMT

←Older revision		Revision as of 12:49, 30 September 2008
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-	The expression is so complicated that we first need to simplify it. We start with the three fractions,	+	The expression is so complicated that we first need to simplify it. We start with the three fractions, <math>1/\!\sqrt{3}</math>, <math>1/\!\sqrt{5}</math> and <math>1/\!\sqrt{2}</math>, which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators
-	<math>\~~frac{1}{~~\sqrt{3}},\~~quad \frac{1}{~~\sqrt{5}}</math>	+
-	and	+
-	<math>~~\frac{~~1}{\sqrt{2}}</math>, which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators:	+

		+	{{Displayed math\|\|<math>\frac{\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}}{\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}}\,\textrm{.}</math>}}

-	<math>\frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{1}{\sqrt{3}}\centerdot \frac{\sqrt{3}}{\sqrt{3}}-\frac{1}{\sqrt{5}}\centerdot \frac{\sqrt{5}}{\sqrt{5}}}{\frac{1}{\sqrt{2}}\centerdot \frac{\sqrt{2}}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5}}{\frac{\sqrt{2}}{2}-\frac{1}{2}}</math>	+	Then, multiply the top and bottom of the fraction by 2 so that we get rid of the fractions in the denominator
-		+
-		+
-	Then, multiply the top and bottom of the fraction by	+
-	~~<math>\text{~~2~~}</math>~~	+
-	so that we get rid of the fractions in the denominator:	+
-		+
-		+
-	<math>\frac{\left( \frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5} \right)\centerdot 2}{\left( \frac{\sqrt{2}}{2}-\frac{1}{2} \right)\centerdot 2}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\frac{2\sqrt{2}}{2}-\frac{2}{2}}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}</math>	+

		+	{{Displayed math\|\|<math>\frac{\Bigl(\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}\Bigr)\cdot 2}{\Bigl(\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\Bigr)\cdot 2} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\dfrac{2\sqrt{2}}{2}-\dfrac{2}{2}} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\,\textrm{.}</math>}}

	Now, we can multiply the top and bottom by the conjugate of the denominator		Now, we can multiply the top and bottom by the conjugate of the denominator
	<math>\sqrt{2}+1</math>, to get an expression without roots in the denominator.		<math>\sqrt{2}+1</math>, to get an expression without roots in the denominator.

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}
-	& \frac{\~~frac~~{2\sqrt{3}}{3}-\~~frac~~{2\sqrt{5}}{5}}{\sqrt{2}-1}=\frac{\~~frac~~{2\sqrt{3}}{3}-\~~frac~~{2\sqrt{5}}{5}}{\sqrt{2}-1}\~~centerdot~~ \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\~~left~~( \~~frac~~{2\sqrt{3}}{3}-\~~frac~~{2\sqrt{5}}{5} \~~right~~)~~\left~~( \sqrt{2}+1 ~~\right~~)}{~~\left~~( \sqrt{2} ~~\right~~)^{2}-1^{2}} \\	+	&= \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}\\[10pt]
-	& =\frac{\~~frac~~{2\sqrt{3}\sqrt{2}}{3}+\~~frac~~{2\sqrt{3}\~~centerdot~~ 1}{3}-\~~frac~~{2\sqrt{5}\sqrt{2}}{5}-\~~frac~~{2\sqrt{5}\~~centerdot~~ 1}{5}}{2-1}=\frac{\~~frac~~{2}{3}\sqrt{3\~~centerdot~~ 2}+\~~frac~~{2}{3}\sqrt{3}-\~~frac~~{2}{5}\sqrt{10}-\~~frac~~{2}{5}\sqrt{5}}{1} \\	+	&= \frac{\Bigl(\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}\Bigr)(\sqrt{2}+1)}{(\sqrt{2})^{2}-1^{2}}\\[10pt]
-	& =\frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5} \\	+	&= \frac{\dfrac{2\sqrt{3}\sqrt{2}}{3}+\dfrac{2\sqrt{3}\cdot 1}{3}-\dfrac{2\sqrt{5}\sqrt{2}}{5}-\dfrac{2\sqrt{5}\cdot 1}{5}}{2-1}\\[10pt]
-	\end{align}</math>	+	&= \frac{\dfrac{2}{3}\sqrt{3\cdot 2}+\dfrac{2}{3}\sqrt{3}-\dfrac{2}{5}\sqrt{2\cdot 5}-\dfrac{2}{5}\sqrt{5}}{1}\\[10pt]
		+	&= \frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5}\,\textrm{.}
		+	\end{align}</math>}}

Ian at 09:24, 23 September 2008

Ian — Tue, 23 Sep 2008 09:24:31 GMT

←Older revision		Revision as of 09:24, 23 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The expression is so complicated that we first need to simplify it. We start with the three fractions,
-	<~~center~~> ~~[[Image:3_1_6c-~~1(2~~).gif]]~~ </~~center~~>	+	<math>\frac{1}{\sqrt{3}},\quad \frac{1}{\sqrt{5}}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
-	{{~~NAVCONTENT_START~~}}	+	<math>\frac{1}{\sqrt{2}}</math>, which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators:
-	<~~center~~> ~~[[Image~~:~~3_1_6c~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{1}{\sqrt{3}}\centerdot \frac{\sqrt{3}}{\sqrt{3}}-\frac{1}{\sqrt{5}}\centerdot \frac{\sqrt{5}}{\sqrt{5}}}{\frac{1}{\sqrt{2}}\centerdot \frac{\sqrt{2}}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5}}{\frac{\sqrt{2}}{2}-\frac{1}{2}}</math>
		+
		+
		+	Then, multiply the top and bottom of the fraction by
		+	<math>\text{2}</math>
		+	so that we get rid of the fractions in the denominator:
		+
		+
		+	<math>\frac{\left( \frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5} \right)\centerdot 2}{\left( \frac{\sqrt{2}}{2}-\frac{1}{2} \right)\centerdot 2}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\frac{2\sqrt{2}}{2}-\frac{2}{2}}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}</math>
		+
		+
		+	Now, we can multiply the top and bottom by the conjugate of the denominator
		+	<math>\sqrt{2}+1</math>, to get an expression without roots in the denominator.
		+
		+
		+	<math>\begin{align}
		+	& \frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}\centerdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left( \frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5} \right)\left( \sqrt{2}+1 \right)}{\left( \sqrt{2} \right)^{2}-1^{2}} \\
		+	& =\frac{\frac{2\sqrt{3}\sqrt{2}}{3}+\frac{2\sqrt{3}\centerdot 1}{3}-\frac{2\sqrt{5}\sqrt{2}}{5}-\frac{2\sqrt{5}\centerdot 1}{5}}{2-1}=\frac{\frac{2}{3}\sqrt{3\centerdot 2}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5}}{1} \\
		+	& =\frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5} \\
		+	\end{align}</math>

Tekbot: Lösning 3.1:6c moved to Solution 3.1:6c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:14:23 GMT

Lösning 3.1:6c moved to Solution 3.1:6c: Robot: moved page

←Older revision

Revision as of 09:14, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:54:19 GMT

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

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Oskar — Tue, 01 Apr 2008 14:10:18 GMT

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