Lösung 3.1:5c

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The trick is to use the conjugate rule
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<center> [[Image:3_1_5c.gif]] </center>
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<math>\left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}}</math>
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and multiply the top and bottom of the fraction by
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<math>3-\sqrt{7}</math>
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(note the minus sign), since then the new denominator will be
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<math>\left( 3+\sqrt{7} \right)\left( 3-\sqrt{7} \right)=3^{2}-\left( \sqrt{7} \right)^{2}=9-7=2</math>
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(conjugate rule with
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<math>a=\text{3 }</math>
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and
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<math>b=\sqrt{\text{7}}</math>
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), i.e. the root sign is squared away.
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The whole calculation is
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<math>\begin{align}
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& \frac{2}{3+\sqrt{7}}=\frac{2}{3+\sqrt{7}}\centerdot \frac{3-\sqrt{7}}{3-\sqrt{7}}=\frac{2\left( 3-\sqrt{7} \right)}{3^{2}-\left( \sqrt{7} \right)^{2}} \\
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& =\frac{2\centerdot 3-2\sqrt{7}}{2}=3-\sqrt{7} \\
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\end{align}</math>

Version vom 14:43, 22. Sep. 2008

The trick is to use the conjugate rule \displaystyle \left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}} and multiply the top and bottom of the fraction by \displaystyle 3-\sqrt{7} (note the minus sign), since then the new denominator will be \displaystyle \left( 3+\sqrt{7} \right)\left( 3-\sqrt{7} \right)=3^{2}-\left( \sqrt{7} \right)^{2}=9-7=2 (conjugate rule with \displaystyle a=\text{3 } and \displaystyle b=\sqrt{\text{7}} ), i.e. the root sign is squared away.

The whole calculation is


\displaystyle \begin{align} & \frac{2}{3+\sqrt{7}}=\frac{2}{3+\sqrt{7}}\centerdot \frac{3-\sqrt{7}}{3-\sqrt{7}}=\frac{2\left( 3-\sqrt{7} \right)}{3^{2}-\left( \sqrt{7} \right)^{2}} \\ & =\frac{2\centerdot 3-2\sqrt{7}}{2}=3-\sqrt{7} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:5c“