Lösung 3.1:6c

Aus Online Mathematik Brückenkurs 1

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Aktuelle Version (09:57, 10. Jun. 2009) (bearbeiten) (rückgängig)
 
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The expression is so complicated that we first need to simplify it. We start with the three fractions, <math>1/\!\sqrt{3}</math>, <math>1/\!\sqrt{5}</math> and <math>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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Zuerst schreiben wir die Brüche <math>1/\!\sqrt{3}</math>, <math>1/\!\sqrt{5}</math> und <math>1/\!\sqrt{2}</math>, sodass sie nur Wurzeln im Zähler enthalten
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{{Displayed math||<math>\frac{\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}}{\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}}{\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}}\,\textrm{.}</math>}}
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Then, multiply the top and bottom of the fraction by 2 so that we get rid of the fractions in the denominator
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Wir erweitern den gesamten Bruch mit 2, um die Nenner im Nenner los zu werden
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{{Displayed math||<math>\frac{\Bigl(\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}\Bigr)\cdot 2}{\Bigl(\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\Bigr)\cdot 2} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\dfrac{2\sqrt{2}}{2}-\dfrac{2}{2}} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{\Bigl(\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}\Bigr)\cdot 2}{\Bigl(\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\Bigr)\cdot 2} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\dfrac{2\sqrt{2}}{2}-\dfrac{2}{2}} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\,\textrm{.}</math>}}
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Now, we can multiply the top and bottom by the conjugate of the denominator
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Jetzt erweitern wir den ganzen Bruch mit den konjugierten Nenner <math>\sqrt{2}+1</math> und vereinfachen
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<math>\sqrt{2}+1</math>, to get an expression without roots in the denominator.
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}
\frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}
&= \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}\\[10pt]
&= \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}\\[10pt]

Aktuelle Version

Zuerst schreiben wir die Brüche \displaystyle 1/\!\sqrt{3}, \displaystyle 1/\!\sqrt{5} und \displaystyle 1/\!\sqrt{2}, sodass sie nur Wurzeln im Zähler enthalten

\displaystyle \frac{\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}}{\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}}\,\textrm{.}

Wir erweitern den gesamten Bruch mit 2, um die Nenner im Nenner los zu werden

\displaystyle \frac{\Bigl(\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}\Bigr)\cdot 2}{\Bigl(\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\Bigr)\cdot 2} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\dfrac{2\sqrt{2}}{2}-\dfrac{2}{2}} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\,\textrm{.}

Jetzt erweitern wir den ganzen Bruch mit den konjugierten Nenner \displaystyle \sqrt{2}+1 und vereinfachen

\displaystyle \begin{align}

\frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1} &= \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}\\[10pt] &= \frac{\Bigl(\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}\Bigr)(\sqrt{2}+1)}{(\sqrt{2})^{2}-1^{2}}\\[10pt] &= \frac{\dfrac{2\sqrt{3}\sqrt{2}}{3}+\dfrac{2\sqrt{3}\cdot 1}{3}-\dfrac{2\sqrt{5}\sqrt{2}}{5}-\dfrac{2\sqrt{5}\cdot 1}{5}}{2-1}\\[10pt] &= \frac{\dfrac{2}{3}\sqrt{3\cdot 2}+\dfrac{2}{3}\sqrt{3}-\dfrac{2}{5}\sqrt{2\cdot 5}-\dfrac{2}{5}\sqrt{5}}{1}\\[10pt] &= \frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5}\,\textrm{.} \end{align}

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