Lösung 3.1:6c

Aus Online Mathematik Brückenkurs 1

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K (Lösning 3.1:6c moved to Solution 3.1:6c: Robot: moved page)
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The expression is so complicated that we first need to simplify it. We start with the three fractions,
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<center> [[Image:3_1_6c-1(2).gif]] </center>
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<math>\frac{1}{\sqrt{3}},\quad \frac{1}{\sqrt{5}}</math>
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and
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<math>\frac{1}{\sqrt{2}}</math>, which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators:
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<center> [[Image:3_1_6c-2(2).gif]] </center>
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<math>\frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{1}{\sqrt{3}}\centerdot \frac{\sqrt{3}}{\sqrt{3}}-\frac{1}{\sqrt{5}}\centerdot \frac{\sqrt{5}}{\sqrt{5}}}{\frac{1}{\sqrt{2}}\centerdot \frac{\sqrt{2}}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5}}{\frac{\sqrt{2}}{2}-\frac{1}{2}}</math>
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Then, multiply the top and bottom of the fraction by
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<math>\text{2}</math>
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so that we get rid of the fractions in the denominator:
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<math>\frac{\left( \frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5} \right)\centerdot 2}{\left( \frac{\sqrt{2}}{2}-\frac{1}{2} \right)\centerdot 2}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\frac{2\sqrt{2}}{2}-\frac{2}{2}}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}</math>
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Now, we can multiply the top and bottom by the conjugate of the denominator
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<math>\sqrt{2}+1</math>, to get an expression without roots in the denominator.
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<math>\begin{align}
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& \frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}\centerdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left( \frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5} \right)\left( \sqrt{2}+1 \right)}{\left( \sqrt{2} \right)^{2}-1^{2}} \\
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& =\frac{\frac{2\sqrt{3}\sqrt{2}}{3}+\frac{2\sqrt{3}\centerdot 1}{3}-\frac{2\sqrt{5}\sqrt{2}}{5}-\frac{2\sqrt{5}\centerdot 1}{5}}{2-1}=\frac{\frac{2}{3}\sqrt{3\centerdot 2}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5}}{1} \\
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& =\frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5} \\
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\end{align}</math>

Version vom 09:24, 23. Sep. 2008

The expression is so complicated that we first need to simplify it. We start with the three fractions, \displaystyle \frac{1}{\sqrt{3}},\quad \frac{1}{\sqrt{5}} and \displaystyle \frac{1}{\sqrt{2}}, which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators:


\displaystyle \frac{\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{1}{\sqrt{3}}\centerdot \frac{\sqrt{3}}{\sqrt{3}}-\frac{1}{\sqrt{5}}\centerdot \frac{\sqrt{5}}{\sqrt{5}}}{\frac{1}{\sqrt{2}}\centerdot \frac{\sqrt{2}}{\sqrt{2}}-\frac{1}{2}}=\frac{\frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5}}{\frac{\sqrt{2}}{2}-\frac{1}{2}}


Then, multiply the top and bottom of the fraction by \displaystyle \text{2} so that we get rid of the fractions in the denominator:


\displaystyle \frac{\left( \frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{5} \right)\centerdot 2}{\left( \frac{\sqrt{2}}{2}-\frac{1}{2} \right)\centerdot 2}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\frac{2\sqrt{2}}{2}-\frac{2}{2}}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}


Now, we can multiply the top and bottom by the conjugate of the denominator \displaystyle \sqrt{2}+1, to get an expression without roots in the denominator.


\displaystyle \begin{align} & \frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}=\frac{\frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5}}{\sqrt{2}-1}\centerdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left( \frac{2\sqrt{3}}{3}-\frac{2\sqrt{5}}{5} \right)\left( \sqrt{2}+1 \right)}{\left( \sqrt{2} \right)^{2}-1^{2}} \\ & =\frac{\frac{2\sqrt{3}\sqrt{2}}{3}+\frac{2\sqrt{3}\centerdot 1}{3}-\frac{2\sqrt{5}\sqrt{2}}{5}-\frac{2\sqrt{5}\centerdot 1}{5}}{2-1}=\frac{\frac{2}{3}\sqrt{3\centerdot 2}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5}}{1} \\ & =\frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:6c“