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Lösung 3.4:1b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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In our example, we write first
In our example, we write first
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{{Displayed math||<math>\frac{x^2}{x+1}=\frac{x^2+x-x}{x+1}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{x^2}{x+1}=\frac{x^2+x-x}{x+1}\,\textrm{.}</math>}}
Note that we have added <math>x</math> and then taken away the same <math>x</math> in the numerator. The reason why we do it in this way is that we can now eliminate terms in <math>x^2</math>,
Note that we have added <math>x</math> and then taken away the same <math>x</math> in the numerator. The reason why we do it in this way is that we can now eliminate terms in <math>x^2</math>,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{x^2+x-x}{x+1}
\frac{x^2+x-x}{x+1}
&= \frac{x^2+x}{x+1}-\frac{x}{x+1}
&= \frac{x^2+x}{x+1}-\frac{x}{x+1}
Zeile 18: Zeile 18:
We can treat the second term <math>x/(x+1)</math> in a similar way. Add and take away <math>1</math> in order to get <math>x+1</math>, which can then be divided away,
We can treat the second term <math>x/(x+1)</math> in a similar way. Add and take away <math>1</math> in order to get <math>x+1</math>, which can then be divided away,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
x-\frac{x}{x+1} = x-\frac{x+1-1}{x+1} = x-\frac{x+1}{x+1}+\frac{1}{x+1}
x-\frac{x}{x+1} = x-\frac{x+1-1}{x+1} = x-\frac{x+1}{x+1}+\frac{1}{x+1}
= x-1+\frac{1}{x+1}\,\textrm{.}
= x-1+\frac{1}{x+1}\,\textrm{.}

Version vom 13:14, 10. Mär. 2009

The trick is to complement the numerator with terms that are missing so that the division of the leading term (the term of highest degree) becomes possible, and then successively work down in degree until the numerator has lower degree than the denominator.

In our example, we write first

x2x+1=x+1x2+x−x.

Note that we have added x and then taken away the same x in the numerator. The reason why we do it in this way is that we can now eliminate terms in x2,

x+1x2+x−x=x+1x2+x−xx+1=x+1x(x+1)−xx+1=x−xx+1.

Thus, we added x which is precisely what is necessary in order that the numerator x2 should contain a sub expression which has x+1 as a factor.

We can treat the second term x(x+1) in a similar way. Add and take away 1 in order to get x+1, which can then be divided away,

x−xx+1=x−x+1x+1−1=x−x+1x+1+1x+1=x−1+1x+1.

In the remaining polynomial fraction, the numerator is of lower degree than the denominator and we cannot go any further with the division.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1b“