løs likninga

[Janhaa]

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

$(\frac{1}{x}+1{)}^{x+1}=(\frac{1}{2020}+1{)}^{2020}$

[Mattebruker]

( $\frac{1}{x}$ + 1 )${}^{x+1}$ = ( $\frac{1}{x}$ + 1 )${}^x$ $\cdot$( $\frac{1}{x}$ + 1)


( $\frac{1}{x}$ + 1 )${}^x$ $\cdot$( $\frac{1}{x}$ + 1) = ( $\frac{1}{2020}$ + 1 )${}^{2020}$

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

[Janhaa]

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

$(\frac{1}{x}+1{)}^{x+1}=(\frac{1}{2020}+1{)}^{2020}$

Løste den slik:

x = -2021:

$$\begin{gathered} (\frac{x+1}{x}{)}^{x+1}=(\frac{2021}{2020}{)}^{2020} \\ (\frac{x+1}{x}{)}^{x+1}=(\frac{2020}{2021}{)}^{-2020} \\ (\frac{x+1}{x}{)}^{x+1}=(\frac{-2020}{-2021}{)}^{-2020} \\ dvs:x=-2021 \end{gathered}$$
og
$$\begin{gathered} x+1=-2020 \\ x=-2021 \end{gathered}$$

[Mattebruker]

Kreativ løysing !