Lösung 3.1:3d

Aus Online Mathematik Brückenkurs 1

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We can multiply
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<center> [[Image:3_1_3d.gif]] </center>
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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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<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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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>

Version vom 13:07, 22. Sep. 2008

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


\displaystyle \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}}.

Because \displaystyle \frac{2\centerdot 6}{3}=2\centerdot 2=2^{2} and \displaystyle \frac{2\centerdot 3}{3}=2, we obtain


\displaystyle \sqrt{\frac{2}{3}}\left( \sqrt{6}-\sqrt{3} \right)=\sqrt{2^{2}}-\sqrt{2}=2-\sqrt{2}

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