Solution 2.1:6d

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m (Lösning 2.1:6d moved to Solution 2.1:6d: Robot: moved page)
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First, we simplify the numerator and denominator for the whole fraction by rewriting as
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<center> [[Image:2_1_6d.gif]] </center>
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<math>\begin{align}
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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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<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>

Revision as of 11:23, 16 September 2008

First, we simplify the numerator and denominator for the whole fraction by rewriting as


\displaystyle \begin{align} & 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} \\ & \frac{a^{2}-b^{2}+b^{2}}{a+b}=\frac{a^{2}}{a+b} \\ \end{align}


\displaystyle \begin{align} & 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}} \\ & =\frac{\left( a+b \right)^{2}-\left( a-b \right)^{2}}{\left( a+b \right)^{2}} \\ & =\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}} \\ \end{align}


The whole fraction is therefore


\displaystyle \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}

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