Lösung 2.1:1b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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When the factor <math>xy</math> is multiplied by the expression inside the brackets, <math> 1+x+x^2 </math>, the distributive rule gives that all three terms <math>1</math>, <math>x</math> and <math>-x^2</math> are multiplied by <math>xy</math>, | When the factor <math>xy</math> is multiplied by the expression inside the brackets, <math> 1+x+x^2 </math>, the distributive rule gives that all three terms <math>1</math>, <math>x</math> and <math>-x^2</math> are multiplied by <math>xy</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(1+x-x^2) &= 1\cdot xy + x\cdot xy -x^2\cdot xy\\[3pt] | (1+x-x^2) &= 1\cdot xy + x\cdot xy -x^2\cdot xy\\[3pt] | ||
&= xy+x^2y-x^3y\,\textrm{.} | &= xy+x^2y-x^3y\,\textrm{.} | ||
\end{align} | \end{align} | ||
</math>}} | </math>}} | ||
Version vom 08:20, 22. Okt. 2008
When the factor \displaystyle xy is multiplied by the expression inside the brackets, \displaystyle 1+x+x^2 , the distributive rule gives that all three terms \displaystyle 1, \displaystyle x and \displaystyle -x^2 are multiplied by \displaystyle xy,
| \displaystyle \begin{align}
(1+x-x^2) &= 1\cdot xy + x\cdot xy -x^2\cdot xy\\[3pt] &= xy+x^2y-x^3y\,\textrm{.} \end{align} |
