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Solution 2.1:7b

From Förberedande kurs i matematik 1

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Current revision (12:00, 23 September 2008) (edit) (undo)
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The denominators
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The denominators <math>x-1</math> and <math>x^{2}</math> do not have a common denominator, so the lowest common denominator is <math>x^{2}(x-1)</math>. We treat all three terms so that they have a common denominator and then start simplifying
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<math>x-1</math>
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and
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<math>x^{2}</math>
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do not have a common denominator, so the lowest common denominator is
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<math>x^{2}\left( x-1 \right)</math>. We treat all three terms so that they have a common denominator and then start simplifying:
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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x + \frac{1}{x-1} + \frac{1}{x^{2}}
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& x+\frac{1}{x-1}+\frac{1}{x^{2}}=x\centerdot \frac{x^{2}\left( x-1 \right)}{x^{2}\left( x-1 \right)}+\frac{1}{x-1}\centerdot \frac{x^{2}}{x^{2}}+\frac{1}{x^{2}}\centerdot \frac{x-1}{x-1} \\
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&= x\cdot\frac{x^{2}(x-1)}{x^{2}(x-1)} + \frac{1}{x-1}\cdot\frac{x^{2}}{x^{2}} + \frac{1}{x^{2}}\cdot\frac{x-1}{x-1}\\[5pt]
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& =\frac{x^{3}\left( x-1 \right)+x^{2}+\left( x-1 \right)}{x^{2}\left( x-1 \right)}=\frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}\left( x-1 \right)} \\
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&= \frac{x^{3}(x-1)+x^{2}+(x-1)}{x^{2}(x-1)}\\[5pt]
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\end{align}</math>
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&= \frac{x^{4}-x^{3}+x^{2}+x-1}{x^{2}(x-1)}\,\textrm{.}
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\end{align}</math>}}

Current revision

The denominators x1 and x2 do not have a common denominator, so the lowest common denominator is x2(x1). We treat all three terms so that they have a common denominator and then start simplifying

x+1x1+1x2=xx2(x1)x2(x1)+1x1x2x2+1x2x1x1=x2(x1)x3(x1)+x2+(x1)=x2(x1)x4x3+x2+x1.