Lösung 2.1:5c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
The fraction can be further simplified if it is possible to factorize and eliminate common factors from the numerator and denominator. Both numerator and denominator are already factorized to a certain extent, but we can go further with the numerator and break it up into linear factors by using the conjugate rule
The fraction can be further simplified if it is possible to factorize and eliminate common factors from the numerator and denominator. Both numerator and denominator are already factorized to a certain extent, but we can go further with the numerator and break it up into linear factors by using the conjugate rule
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
3x^{2}-12 &= 3(x^{2}-4) = 3(x+2)(x-2)\,,\\
3x^{2}-12 &= 3(x^{2}-4) = 3(x+2)(x-2)\,,\\
x^{2}-1 &= (x+1)(x-1) \,\textrm{.}
x^{2}-1 &= (x+1)(x-1) \,\textrm{.}
Zeile 8: Zeile 8:
The whole expression is therefore equal to
The whole expression is therefore equal to
-
{{Displayed math||<math>\frac{3(x+2)(x-2)(x+1)(x-1)}{(x+1)(x+2)} = 3(x-2)(x-1)\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{3(x+2)(x-2)(x+1)(x-1)}{(x+1)(x+2)} = 3(x-2)(x-1)\,\textrm{.}</math>}}
Note: One can of course expand the expression to get <math>3x^{2}-9x+6</math>
Note: One can of course expand the expression to get <math>3x^{2}-9x+6</math>
as the answer.
as the answer.

Version vom 08:24, 22. Okt. 2008

The fraction can be further simplified if it is possible to factorize and eliminate common factors from the numerator and denominator. Both numerator and denominator are already factorized to a certain extent, but we can go further with the numerator and break it up into linear factors by using the conjugate rule

\displaystyle \begin{align}

3x^{2}-12 &= 3(x^{2}-4) = 3(x+2)(x-2)\,,\\ x^{2}-1 &= (x+1)(x-1) \,\textrm{.} \end{align}

The whole expression is therefore equal to

\displaystyle \frac{3(x+2)(x-2)(x+1)(x-1)}{(x+1)(x+2)} = 3(x-2)(x-1)\,\textrm{.}

Note: One can of course expand the expression to get \displaystyle 3x^{2}-9x+6 as the answer.

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