Lösung 2.1:5b

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We can factorize the denominators as
We can factorize the denominators as
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{{Displayed math||<math>\begin{align}
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y^{2}-2y &= y(y-2)\\
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y^{2}-4 &= (y-2)(y+2)\quad\text{[difference of two squares]}
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\end{align}</math>}}
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<math>y^{2}-2y=y\left( y-2 \right)</math>
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and then we see that the terms' lowest common denominator is <math>y(y-2)(y+2)</math> because it is the product that contains the smallest number of factors which contain both <math>y(y-2)</math> and <math>(y-2)(y+2)</math>.
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Now, we rewrite the fractions so that they have same denominators and start simplifying
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{{Displayed math||<math>\begin{align}
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\frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}
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&= \frac{1}{y(y-2)}\cdot\frac{y+2}{y+2}-\frac{2}{(y-2)(y+2)}\cdot\frac{y}{y}\\[5pt]
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&= \frac{y+2}{y(y-2)(y+2)} - \frac{2y}{(y-2)(y+2)y}\\[5pt]
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&= \frac{y+2-2y}{y(y-2)(y+2)}\\[5pt]
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&= \frac{-y+2}{y(y-2)(y+2)}\,\textrm{.}
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\end{align}</math>}}
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The numerator can be rewritten as <math>-y+2=-(y-2)</math> and we can eliminate the common factor <math>y-2</math>,
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<math>y^{2}-4=\left( y-2 \right)\left( y+2 \right)</math>
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{{Displayed math||<math>\frac{-y+2}{y(y-2)(y+2)} = \frac{-(y-2)}{y(y-2)(y+2)} = \frac{-1}{y(y+2)} = -\frac{1}{y(y+2)}\,\textrm{.}</math>}}
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[by the conjugate rule]
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and then we see that the terms' lowest common denominator is
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<math>y\left( y-2 \right)\left( y+2 \right)</math>
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because it is the product that contains the smallest number of factors which contain both
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<math>y\left( y-2 \right)</math>
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and
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<math>\left( y-2 \right)\left( y+2 \right)</math>
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.
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Now, we rewrite the fractions so that they have same denominators and start simplifying:
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<math>\begin{align}
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& y^{2}-4=\left( y-2 \right)\left( y+2 \right) \\
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& y^{2}-2y=y\left( y-2 \right) \\
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& \frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}=\frac{1}{y\left( y-2 \right)}\centerdot \frac{y+2}{y+2}-\frac{2}{\left( y-2 \right)\left( y+2 \right)}\centerdot \frac{y}{y} \\
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& \\
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& =\frac{y+2}{y\left( y-2 \right)\left( y+2 \right)}-\frac{2y}{\left( y-2 \right)\left( y+2 \right)y} \\
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& \\
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& =\frac{y+2-2y}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)} \\
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\end{align}</math>
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The numerator can be rewritten as
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<math>-y+2=-\left( y-2 \right)</math>
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and we can eliminate the common factor
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<math>y-2</math>
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.
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<math>\frac{-y+2}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-\left( y-2 \right)}{y\left( y-2 \right)\left( y+2 \right)}=\frac{-1}{y\left( y+2 \right)}=-\frac{1}{y\left( y+2 \right)}</math>
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Version vom 10:51, 23. Sep. 2008

We can factorize the denominators as

Vorlage:Displayed math

and then we see that the terms' lowest common denominator is \displaystyle y(y-2)(y+2) because it is the product that contains the smallest number of factors which contain both \displaystyle y(y-2) and \displaystyle (y-2)(y+2).

Now, we rewrite the fractions so that they have same denominators and start simplifying

Vorlage:Displayed math

The numerator can be rewritten as \displaystyle -y+2=-(y-2) and we can eliminate the common factor \displaystyle y-2,

Vorlage:Displayed math

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