Lösung 2.1:5a

Aus Online Mathematik Brückenkurs 1

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In the same way that we calculated fractions, we can subtract the terms' numerators if we first
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<center> [[Image:2_1_5a.gif]] </center>
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expand the fractions so that they have the same denominator. Because the denominators are
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<math>x-x^{2}=x\left( 1-x \right)</math>
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and
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<math>x</math>, the lowest common denominator is
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<math>x\left( 1-x \right)</math>:
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<math>\begin{align}
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& \frac{1}{x-x^{2}}-\frac{1}{x}=\frac{1}{x-x^{2}}-\frac{1}{x}\centerdot \frac{1-x}{1-x}=\frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}} \\
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& \\
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& =\frac{1-\left( 1-x \right)}{x-x^{2}}=\frac{1-1+x}{x-x^{2}}=\frac{x}{x-x^{2}} \\
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& \\
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\end{align}</math>
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This fraction can be simplified by eliminating the factor
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<math>x</math>
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from the numerator and denominator.
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<math>\begin{align}
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& \frac{x}{x-x^{2}}=\frac{x}{x\left( 1-x \right)}=\frac{1}{1-x}. \\
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& \\
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\end{align}</math>

Version vom 14:46, 15. Sep. 2008

In the same way that we calculated fractions, we can subtract the terms' numerators if we first expand the fractions so that they have the same denominator. Because the denominators are \displaystyle x-x^{2}=x\left( 1-x \right) and \displaystyle x, the lowest common denominator is \displaystyle x\left( 1-x \right):


\displaystyle \begin{align} & \frac{1}{x-x^{2}}-\frac{1}{x}=\frac{1}{x-x^{2}}-\frac{1}{x}\centerdot \frac{1-x}{1-x}=\frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}} \\ & \\ & =\frac{1-\left( 1-x \right)}{x-x^{2}}=\frac{1-1+x}{x-x^{2}}=\frac{x}{x-x^{2}} \\ & \\ \end{align}


This fraction can be simplified by eliminating the factor \displaystyle x from the numerator and denominator.


\displaystyle \begin{align} & \frac{x}{x-x^{2}}=\frac{x}{x\left( 1-x \right)}=\frac{1}{1-x}. \\ & \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:5a“