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Løs likninga under, uten bruk Wolfram A. (vgs);

[tex]\large(\frac{1}{x}+1)^{x+1}=(\frac{1}{2020}+1)^{2020}[/tex]

( [tex]\frac{1}{x}[/tex] + 1 )[tex]^{x+1 }[/tex] = ( [tex]\frac{1}{x}[/tex] + 1 )[tex]^{x}[/tex] [tex]\cdot[/tex]( [tex]\frac{1}{x}[/tex] + 1)


( [tex]\frac{1}{x}[/tex] + 1 )[tex]^{x}[/tex] [tex]\cdot[/tex]( [tex]\frac{1}{x}[/tex] + 1) = ( [tex]\frac{1}{2020}[/tex] + 1 )[tex]^{2020}[/tex]

[tex]\Rightarrow[/tex] x [tex]\simeq[/tex] 2020 ettersom "ekstra faktor" på V.S. ( [tex]\frac{1}{2020}[/tex] + 1 ) = 1.0004905 [tex]\simeq[/tex] 1.0

Janhaa skrev:Løs likninga under, uten bruk Wolfram A. (vgs);

[tex]\large(\frac{1}{x}+1)^{x+1}=(\frac{1}{2020}+1)^{2020}[/tex]

Løste den slik:

x = -2021:

[tex](\frac{x+1}{x})^{x+1}=(\frac{2021}{2020})^{2020}\\\\\\(\frac{x+1}{x})^{x+1}=(\frac{2020}{2021})^{-2020}\\\\(\frac{x+1}{x})^{x+1}=(\frac{-2020}{-2021})^{-2020}\\ \\ dvs:\ x=-2021\\[/tex]
og
[tex]x+1=-2020\\ x=-2021[/tex]

Kreativ løysing !