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a fost răspuns

Cum se rezolva n= 5^n x 4^n+1 + 5^n+1x 4n + 20^n+1?

Răspuns :

Renatemambouko
5^nx4^n+1+5^n+1x4^n+20^n+1=
5^nx4^nx4+5^nx5x4^n+20^nx20=
20^nx4+20^nx5+20^nx20=
20^n(4+5+20)=20^n x 29
0000000
[tex]n=5^n*4^{n+1}+5^{n+1}*4^n+20^{n+1} \\ n=5^n*4^{n+1}+5^{n+1}*4^n+5^{n+1}*4^{n+1} \\ n=5^n*4^n(4+5+5*4) \\ n=18*29^n \\ [/tex]
n de dinainte de egal este acelasi cu exponentul de dupa egal ?

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