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[tex]\frac{r+2}{r+3}-\frac{r+2}{2r+1}=\frac{(r+2)\cdot(2r+1)-(r+2)\cdot(r+3)}{(r+3)\cdot(2r+1)}=\frac{(2r^2+5r+2)-(r^2+5r+6)}{2r^2+7r+3}=\frac{r^2-4}{2r^2+7r+3}[/tex]
Tusen takk for hjelpenKnuta skrev:[tex]\frac{r+2}{r+3}-\frac{r+2}{2r+1}=\frac{(r+2)\cdot(2r+1)-(r+2)\cdot(r+3)}{(r+3)\cdot(2r+1)}=\frac{(2r^2+5r+2)-(r^2+5r+6)}{2r^2+7r+3}=\frac{r^2-4}{2r^2+7r+3}[/tex]
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Jeg vet. Jeg lever ikke opp til navnet mitt. ![Embarassed :oops:](./images/smilies/icon_redface.gif)
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