Lösung 2.1:6d

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Aktuelle Version (10:32, 1. Mär. 2009) (bearbeiten) (rückgängig)
 
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First, we simplify the numerator and denominator for the whole fraction by rewriting as
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Zuerst vereinfachen wir jeweils den Zähler und den Nenner des Hauptbruches
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{{Abgesetzte Formel||<math>\begin{align}
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a-b+\frac{b^{2}}{a+b}
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&= (a-b)\cdot\frac{a+b}{a+b} + \frac{b^{2}}{a+b} = \frac{(a-b)\cdot (a+b)+b^{2}}{a+b}\\[5pt]
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&= \frac{a^{2}-b^{2}+b^{2}}{a+b} = \frac{a^{2}}{a+b}\,,\\[15pt]
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1-\biggl(\frac{a-b}{a+b}\biggr)^{2}
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&= 1-\frac{(a-b)^{2}}{(a+b)^{2}} = \frac{(a+b)^{2}}{(a+b)^{2}} - \frac{(a-b)^{2}}{(a+b)^{2}}\\[5pt]
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&= \frac{(a+b)^{2}-(a-b)^{2}}{(a+b)^{2}}\\[5pt]
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&= \frac{(a^{2}+2ab+b^{2})-(a^{2}-2ab+b^{2})}{(a+b)^{2}} = \frac{4ab}{(a+b)^{2}}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Der ganze Bruch ist also
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& a-b+\frac{b^{2}}{a+b}=\left( a-b \right)\centerdot \frac{a+b}{a+b}+\frac{b^{2}}{a+b}=\frac{\left( a-b \right)\centerdot \left( a+b \right)+b^{2}}{a+b} \\
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& \frac{a^{2}-b^{2}+b^{2}}{a+b}=\frac{a^{2}}{a+b} \\
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\end{align}</math>
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{{Abgesetzte Formel||<math>\frac{a-b+\dfrac{b^{2}}{a+b}}{1-\biggl(\dfrac{a-b}{a+b}\biggr)^{2}} = \frac{\dfrac{a^{2}}{a+b}}{\dfrac{4ab}{(a+b)^{2}}} = \frac{a^{2}}{a+b}\cdot\frac{(a+b)^{2}}{4ab} = \frac{a(a+b)}{4b}\,\textrm{.}</math>}}
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<math>\begin{align}
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& 1-\left( \frac{a-b}{a+b} \right)^{2}=1-\frac{\left( a-b \right)^{2}}{\left( a+b \right)^{2}}=\frac{\left( a+b \right)^{2}}{\left( a+b \right)^{2}}-\frac{\left( a-b \right)^{2}}{\left( a+b \right)^{2}} \\
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& =\frac{\left( a+b \right)^{2}-\left( a-b \right)^{2}}{\left( a+b \right)^{2}} \\
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& =\frac{\left( a^{2}+2ab+b^{2} \right)-\left( a^{2}-2ab+b^{2} \right)}{\left( a+b \right)^{2}}=\frac{4ab}{\left( a+b \right)^{2}} \\
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\end{align}</math>
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The whole fraction is therefore
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<math>\frac{a-b+\frac{b^{2}}{a+b}}{1-\left( \frac{a-b}{a+b} \right)^{2}}=\frac{\frac{a^{2}}{a+b}}{\frac{4ab}{\left( a+b \right)^{2}}}=\frac{a^{2}}{a+b}\centerdot \frac{\left( a+b \right)^{2}}{4ab}=\frac{a\left( a+b \right)}{4b}</math>
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Aktuelle Version

Zuerst vereinfachen wir jeweils den Zähler und den Nenner des Hauptbruches

\displaystyle \begin{align}

a-b+\frac{b^{2}}{a+b} &= (a-b)\cdot\frac{a+b}{a+b} + \frac{b^{2}}{a+b} = \frac{(a-b)\cdot (a+b)+b^{2}}{a+b}\\[5pt] &= \frac{a^{2}-b^{2}+b^{2}}{a+b} = \frac{a^{2}}{a+b}\,,\\[15pt] 1-\biggl(\frac{a-b}{a+b}\biggr)^{2} &= 1-\frac{(a-b)^{2}}{(a+b)^{2}} = \frac{(a+b)^{2}}{(a+b)^{2}} - \frac{(a-b)^{2}}{(a+b)^{2}}\\[5pt] &= \frac{(a+b)^{2}-(a-b)^{2}}{(a+b)^{2}}\\[5pt] &= \frac{(a^{2}+2ab+b^{2})-(a^{2}-2ab+b^{2})}{(a+b)^{2}} = \frac{4ab}{(a+b)^{2}}\,\textrm{.} \end{align}

Der ganze Bruch ist also

\displaystyle \frac{a-b+\dfrac{b^{2}}{a+b}}{1-\biggl(\dfrac{a-b}{a+b}\biggr)^{2}} = \frac{\dfrac{a^{2}}{a+b}}{\dfrac{4ab}{(a+b)^{2}}} = \frac{a^{2}}{a+b}\cdot\frac{(a+b)^{2}}{4ab} = \frac{a(a+b)}{4b}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:6d“