Lösung 2.1:1d

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After <math> x^3y^2 </math> are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator,
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After <math> x^3y^2 </math> are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator.
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<math> \qquad \begin{align}
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{{Displayed math||<math>\begin{align}
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x^3y^2\Big( \frac 1y - \frac 1{xy} +1 \Big) &= x^3y^2 \cdot\frac 1y -x^3y^2 \cdot \frac 1{xy} +x^3y^2\cdot 1 \\
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x^3y^2\Big( \frac{1}{y} - \frac{1}{xy} +1 \Big) &= x^3y^2 \cdot\frac{1}{y} -x^3y^2 \cdot \frac{1}{xy} +x^3y^2\cdot 1 \\
&=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\
&=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\
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&=x^3y - x^2y +x^3y^2
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&=x^3y - x^2y +x^3y^2\,,
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\end{align}</math>
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\end{align}</math>}}
where we have used
where we have used
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<math> \qquad \frac{x^3y^2}{y}= \frac{x^3\cdot y\cdot y}{y}= x^3y </math>,
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{{Displayed math||<math>\begin{align}
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\frac{x^3y^2}{y} &= \frac{x^3\cdot y\cdot{}\rlap{/}y}{\rlap{/}y}= x^3y\,,\\[5pt]
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<math>\qquad \frac{x^3y^2}{xy}=\frac{x\cdot x\cdot x \cdot y \cdot y}{x\cdot y} = x\cdot x\cdot y = x^2y </math>
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\frac{x^3y^2}{xy} &= \frac{\rlap{/}x\cdot x\cdot x \cdot y \cdot {}\rlap{/}y}{\rlap{/}x\cdot {}\rlap{/}y} = x\cdot x\cdot y = x^2y\,\textrm{.}\end{align}</math>}}
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Version vom 07:48, 23. Sep. 2008

After \displaystyle x^3y^2 are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator,

Vorlage:Displayed math

where we have used

Vorlage:Displayed math

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