Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 3.4:1e

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 2: Zeile 2:
<math>x^2+3x+1</math>, we need to add and subtract <math>3x^2+x</math> in order to obtain the expression <math>x^3+3x^2+x=x(x^2+3x+1)</math>,
<math>x^2+3x+1</math>, we need to add and subtract <math>3x^2+x</math> in order to obtain the expression <math>x^3+3x^2+x=x(x^2+3x+1)</math>,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{x^3+2x^2+1}{x^2+3x+1}
\frac{x^3+2x^2+1}{x^2+3x+1}
&= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
&= \frac{x^3\bbox[#FFEEAA;,1.5pt]{{}+3x^2+x-3x^2-x}+2x^2+1}{x^2+3x+1}\\[5pt]
Zeile 12: Zeile 12:
Now, we carry out the same procedure with the new quotient. To the term <math>-x^2</math>, we add and subtract <math>-3x-1</math> and obtain
Now, we carry out the same procedure with the new quotient. To the term <math>-x^2</math>, we add and subtract <math>-3x-1</math> and obtain
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
x + \frac{-x^2-x+1}{x^2+3x+1}
x + \frac{-x^2-x+1}{x^2+3x+1}
&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]
&= x + \frac{-x^2\bbox[#FFEEAA;,1.5pt]{{}-3x-1+3x+1}-x+1}{x^2+3x+1}\\[5pt]

Version vom 13:15, 10. Mär. 2009

Imagine for a moment taking away all the terms in the numerator apart from x3. If we are to make x3 divisible by the denominator x2+3x+1, we need to add and subtract 3x2+x in order to obtain the expression x3+3x2+x=x(x2+3x+1),

x2+3x+1x3+2x2+1=x2+3x+1x3+3x2+x−3x2−x+2x2+1=x2+3x+1x3+3x2+x+x2+3x+1−3x2−x+2x2+1=x2+3x+1x(x2+3x+1)+x2+3x+1−x2−x+1=x+x2+3x+1−x2−x+1.

Now, we carry out the same procedure with the new quotient. To the term −x2, we add and subtract −3x−1 and obtain

x+x2+3x+1−x2−x+1=x+x2+3x+1−x2−3x−1+3x+1−x+1=x+x2+3x+1−x2−3x−1+x2+3x+13x+1−x+1=x−1+2x+2x2+3x+1.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:1e“