Lösung 3.1:3d

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We can multiply
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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>,
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<math>\sqrt{\frac{2}{3}}</math>
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into the bracket and then write the root expressions together under a common root sign using the rule
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<math>\sqrt{a}\centerdot \sqrt{b}=\sqrt{ab}</math>
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{{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>}}
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Because <math>(2\cdot 6)/3 = 2\cdot 2 = 2^2</math> and <math>(2\cdot 3)/3 = 2</math>, we obtain
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<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>
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{{Displayed math||<math>\sqrt{\frac{2}{3}}\bigl(\sqrt{6} - \sqrt{3}\bigr) = \sqrt{2^2}-\sqrt{2} = 2-\sqrt{2}\,\textrm{.}</math>}}
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Because
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<math>\frac{2\centerdot 6}{3}=2\centerdot 2=2^{2}</math>
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and
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<math>\frac{2\centerdot 3}{3}=2</math>, we obtain
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<math>\sqrt{\frac{2}{3}}\left( \sqrt{6}-\sqrt{3} \right)=\sqrt{2^{2}}-\sqrt{2}=2-\sqrt{2}</math>
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Version vom 10:40, 30. Sep. 2008

We can multiply \displaystyle \sqrt{\tfrac{2}{3}} into the bracket and then write the root expressions together under a common root sign using the rule \displaystyle \sqrt{a\vphantom{b}}\cdot \sqrt{b} = \sqrt{ab},

Vorlage:Displayed math

Because \displaystyle (2\cdot 6)/3 = 2\cdot 2 = 2^2 and \displaystyle (2\cdot 3)/3 = 2, we obtain

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:3d“