((x^3)^n * y^n(n-1) * z^(n-1)(n+1)) / (x * (y*z)^1+n^2)
Hvordan regner jeg denne ut?
potensregning
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
For å gjøre oppgaven enklare å forstå:
((x^3)^n * y^(n(n-1))) * z^((n-1)(n+1)) / (x * (y*z)^(1+n^2))
((x^3)^n * y^(n(n-1))) * z^((n-1)(n+1)) / (x * (y*z)^(1+n^2))
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- Over-Guru
- Innlegg: 1685
- Registrert: 03/10-2005 12:09
[(x[sup]3[/sup])[sup]n[/sup] * y[sup]n(n-1)[/sup] * z[sup](n-1)(n+1)[/sup]] / [x*(y*z)[sup]1+n^2[/sup]]
= [x[sup]3n[/sup] * y[sup]n^2-n[/sup] * z[sup]n^2-1[/sup]] / [x[sup]1[/sup] * y[sup]n^2+1[/sup] * z[sup]n^2+1[/sup]]
= x[sup]3n-1[/sup] * y[sup]n^2-n-n^2-1[/sup] * z[sup]n^2-1-n^2-1[/sup]
= x[sup]3n-1[/sup] * y[sup]-n-1[/sup] * z[sup]-2[/sup].
= [x[sup]3n[/sup] * y[sup]n^2-n[/sup] * z[sup]n^2-1[/sup]] / [x[sup]1[/sup] * y[sup]n^2+1[/sup] * z[sup]n^2+1[/sup]]
= x[sup]3n-1[/sup] * y[sup]n^2-n-n^2-1[/sup] * z[sup]n^2-1-n^2-1[/sup]
= x[sup]3n-1[/sup] * y[sup]-n-1[/sup] * z[sup]-2[/sup].