Lösung 3.1:5d

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We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression
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<center> [[Image:3_1_5d.gif]] </center>
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<math>\left( a-b \right)\left( a+b \right)=a^{2}-b^{2}</math>, and use the conjugate rule
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with
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<math>a=\sqrt{17}</math>
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and
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<math>b=\sqrt{13}</math>. Both roots are squared away and we get
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<math>\begin{align}
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& \frac{1}{\sqrt{17}-\sqrt{13}}=\frac{1}{\sqrt{17}-\sqrt{13}}\centerdot \frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}} \\
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& =\frac{\sqrt{17}+\sqrt{13}}{\left( \sqrt{17} \right)^{2}-\left( \sqrt{13} \right)^{2}}=\frac{\sqrt{17}+\sqrt{13}}{17-13}=\frac{\sqrt{17}+\sqrt{13}}{4}. \\
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\end{align}</math>
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This expression cannot be simplified any further because neither
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<math>\text{17}</math>
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nor
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<math>\text{13}</math>
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contain any squares as factors.

Version vom 14:51, 22. Sep. 2008

We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression \displaystyle \left( a-b \right)\left( a+b \right)=a^{2}-b^{2}, and use the conjugate rule


with \displaystyle a=\sqrt{17} and \displaystyle b=\sqrt{13}. Both roots are squared away and we get


\displaystyle \begin{align} & \frac{1}{\sqrt{17}-\sqrt{13}}=\frac{1}{\sqrt{17}-\sqrt{13}}\centerdot \frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}} \\ & =\frac{\sqrt{17}+\sqrt{13}}{\left( \sqrt{17} \right)^{2}-\left( \sqrt{13} \right)^{2}}=\frac{\sqrt{17}+\sqrt{13}}{17-13}=\frac{\sqrt{17}+\sqrt{13}}{4}. \\ \end{align}


This expression cannot be simplified any further because neither \displaystyle \text{17} nor \displaystyle \text{13} contain any squares as factors.

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