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ALEXA1011
a fost răspuns

Aflati pe X:
(2013la0+1+2+2la2+2la3+....+2la100)•X=2la101

Răspuns :

Emy78
(2013^0+1+2+2²+2³+....+2^100)·x=
=(1+1+2+2²+2³+....+2^100)·x=
=(2+2+2²+2³+....+2^100)·x=         ( 2+2+2²=2²+2²=2·2²=2³)
= (2³+2³+....+2^100)·x=
=(2^4+....+2^100)·x=
................................
=(2^100+2^100)·x
=2^101·x

Acum putem scrie
2^101·x=2^101
x=1

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