Potenzgesetz < Klassen 8-10 < Schule < Mathe < Vorhilfe
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Hallo,
habe ich das richtig gelöst?
[mm] (\bruch{-r^2}{s^3t^2})^2 [/mm] + [mm] (\bruch{r}{s^2t})^3 [/mm] = [mm] \bruch{-r^4}{s^6t^4} [/mm] + [mm] \bruch{r^3}{s^6t^3} [/mm] = [mm] \bruch{-r^4*t}{s^6t^4} [/mm] + [mm] \bruch{r^3*t}{s^6t^3} [/mm] = [mm] \bruch{-r^4+r^3*t}{s^6t^3}
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(Antwort) fertig  | | Datum: | 20:14 Do 16.02.2012 | | Autor: | notinX |
Hallo,
> Hallo,
>
> habe ich das richtig gelöst?
>
> [mm](\bruch{-r^2}{s^3t^2})^2[/mm] + [mm](\bruch{r}{s^2t})^3[/mm] =
nein, denn es ist [mm] $\left(\frac{-r^2}{s^3t^2}\right)^2=\frac{(-r^2)^2}{(s^3t^2)^2}=\frac{(-1)^2(r^2)^2}{(s^3)^2(t^2)^2}=\frac{r^4}{s^6t^4}$
[/mm]
> [mm]\bruch{-r^4}{s^6t^4}[/mm] + [mm]\bruch{r^3}{s^6t^3}[/mm] =
> [mm]\bruch{-r^4*t}{s^6t^4}[/mm] + [mm]\bruch{r^3*t}{s^6t^3}[/mm] =
> [mm]\bruch{-r^4+r^3*t}{s^6t^3}[/mm]
Gruß,
notinX
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