Lösung 2.1:6b

Aus Online Mathematik Brückenkurs 1

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The lowest common denominator for the three terms is
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<center> [[Image:2_1_6b.gif]] </center>
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<math>\left( x-2 \right)\left( x+3 \right)</math>
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and we expand each term so that all terms have the same denominator:
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<math>\begin{align}
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& \frac{x}{x-2}+\frac{x}{x+3}-2=\frac{x}{x-2}\centerdot \frac{x+3}{x+3}+\frac{x}{x+3}\centerdot \frac{x-2}{x-2}-2\centerdot \frac{\left( x-2 \right)\left( x+3 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{x\left( x+3 \right)+x\left( x-2 \right)-2\left( x-2 \right)\left( x+3 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{x^{2}+3x+x^{2}-2x-2\left( x^{2}+3x-2x-6 \right)}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{\left( x-2 \right)\left( x+3 \right)} \\
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\end{align}</math>
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Now, collect together the terms in the numerator:
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<math>\begin{align}
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& \frac{x}{x-2}+\frac{x}{x+3}-2=\frac{\left( x^{2}+x^{2}-2x^{2} \right)+\left( 3x-2x-6x+4x \right)+12}{\left( x-2 \right)\left( x+3 \right)} \\
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& =\frac{-x+12}{\left( x-2 \right)\left( x+3 \right)} \\
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\end{align}</math>
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NOTE: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.

Version vom 10:53, 16. Sep. 2008

The lowest common denominator for the three terms is \displaystyle \left( x-2 \right)\left( x+3 \right) and we expand each term so that all terms have the same denominator:


\displaystyle \begin{align} & \frac{x}{x-2}+\frac{x}{x+3}-2=\frac{x}{x-2}\centerdot \frac{x+3}{x+3}+\frac{x}{x+3}\centerdot \frac{x-2}{x-2}-2\centerdot \frac{\left( x-2 \right)\left( x+3 \right)}{\left( x-2 \right)\left( x+3 \right)} \\ & =\frac{x\left( x+3 \right)+x\left( x-2 \right)-2\left( x-2 \right)\left( x+3 \right)}{\left( x-2 \right)\left( x+3 \right)} \\ & =\frac{x^{2}+3x+x^{2}-2x-2\left( x^{2}+3x-2x-6 \right)}{\left( x-2 \right)\left( x+3 \right)} \\ & =\frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{\left( x-2 \right)\left( x+3 \right)} \\ \end{align}


Now, collect together the terms in the numerator:


\displaystyle \begin{align} & \frac{x}{x-2}+\frac{x}{x+3}-2=\frac{\left( x^{2}+x^{2}-2x^{2} \right)+\left( 3x-2x-6x+4x \right)+12}{\left( x-2 \right)\left( x+3 \right)} \\ & =\frac{-x+12}{\left( x-2 \right)\left( x+3 \right)} \\ \end{align}


NOTE: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:6b“