Lösung 3.1:3a

Aus Online Mathematik Brückenkurs 1

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{{NAVCONTENT_START}}
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First expand the expression
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<center> [[Image:3_1_3a.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& \left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right)=\sqrt{5}\centerdot \sqrt{5}+\sqrt{5}\centerdot \sqrt{2}-\sqrt{2}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\
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& =\sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\
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\end{align}</math>
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Because
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<math>\sqrt{5}</math>
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and
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<math>\sqrt{2}</math>
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are defined as those numbers which, when multiplied with themselves give
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<math>\text{5}</math>
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and
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<math>2</math> respectively,
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<math>\sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2}=5-2=3</math>
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NOTE: The expansion of
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<math>\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right)</math>
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can also be done directly with the conjugate rule
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<math>\left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}}</math>
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using
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<math>a=\sqrt{5}</math>
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and
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<math>b=\sqrt{2}</math>.

Version vom 12:33, 22. Sep. 2008

First expand the expression


\displaystyle \begin{align} & \left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right)=\sqrt{5}\centerdot \sqrt{5}+\sqrt{5}\centerdot \sqrt{2}-\sqrt{2}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\ & =\sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\ \end{align}


Because \displaystyle \sqrt{5} and \displaystyle \sqrt{2} are defined as those numbers which, when multiplied with themselves give \displaystyle \text{5} and \displaystyle 2 respectively,


\displaystyle \sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2}=5-2=3


NOTE: The expansion of \displaystyle \left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right) can also be done directly with the conjugate rule \displaystyle \left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}} using \displaystyle a=\sqrt{5} and \displaystyle b=\sqrt{2}.

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