Solution 2.1:1d
From Förberedande kurs i matematik 1
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| - | <center> [[Bild:2_1_1d.gif]] </center> | + | 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} | ||
| + | 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 \\ | ||
| + | &=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\ | ||
| + | &=x^3y - x^2y +x^3y^2 | ||
| + | \end{align}</math> | ||
| + | |||
| + | 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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| + | <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> | ||
| + | <!-- <center> [[Bild:2_1_1d.gif]] </center>--> | ||
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Revision as of 08:57, 13 August 2008
After \displaystyle x^3y^2 are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator.
\displaystyle \qquad \begin{align} 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 \\ &=\frac{x^3y^2}{y} -\frac{x^3y^2}{xy} +x^3y^2 \\ &=x^3y - x^2y +x^3y^2 \end{align}
where we have used
\displaystyle \qquad \frac{x^3y^2}{y}= \frac{x^3\cdot y\cdot y}{y}= x^3y ,
\displaystyle \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
