Solution 2.1:7b

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m (Lösning 2.1:7b moved to Solution 2.1:7b: Robot: moved page)
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The denominators
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<center> [[Image:2_1_7b.gif]] </center>
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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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<math>\begin{align}
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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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& =\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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\end{align}</math>

Revision as of 12:46, 16 September 2008

The denominators \displaystyle x-1 and \displaystyle x^{2} do not have a common denominator, so the lowest common denominator is \displaystyle x^{2}\left( x-1 \right). We treat all three terms so that they have a common denominator and then start simplifying:


\displaystyle \begin{align} & 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} \\ & =\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)} \\ \end{align}

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