Lösung 2.1:5b

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We can factorize the denominators as
We can factorize the denominators as
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
y^{2}-2y &= y(y-2)\\
y^{2}-2y &= y(y-2)\\
y^{2}-4 &= (y-2)(y+2)\quad\text{[difference of two squares]}
y^{2}-4 &= (y-2)(y+2)\quad\text{[difference of two squares]}
Zeile 10: Zeile 10:
Now, we rewrite the fractions so that they have same denominators and start simplifying
Now, we rewrite the fractions so that they have same denominators and start simplifying
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}
\frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4}
&= \frac{1}{y(y-2)}\cdot\frac{y+2}{y+2}-\frac{2}{(y-2)(y+2)}\cdot\frac{y}{y}\\[5pt]
&= \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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The numerator can be rewritten as <math>-y+2=-(y-2)</math> and we can eliminate the common factor <math>y-2</math>,
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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{{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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{{Abgesetzte Formel||<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>}}

Version vom 08:24, 22. Okt. 2008

We can factorize the denominators as

\displaystyle \begin{align}

y^{2}-2y &= y(y-2)\\ y^{2}-4 &= (y-2)(y+2)\quad\text{[difference of two squares]} \end{align}

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

\displaystyle \begin{align}

\frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4} &= \frac{1}{y(y-2)}\cdot\frac{y+2}{y+2}-\frac{2}{(y-2)(y+2)}\cdot\frac{y}{y}\\[5pt] &= \frac{y+2}{y(y-2)(y+2)} - \frac{2y}{(y-2)(y+2)y}\\[5pt] &= \frac{y+2-2y}{y(y-2)(y+2)}\\[5pt] &= \frac{-y+2}{y(y-2)(y+2)}\,\textrm{.} \end{align}

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

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