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Question Number 66396 by hmamarques1994@gmai.com last updated on 14/Aug/19
Seja53log1eπ[(x+11)!9999999]=1.Calculex1x2+0,9.
Answered by mr W last updated on 14/Aug/19
53log1eπ[(x+11)!9999999]=1⇒log1eπ[(x+11)!9999999]=0⇒(x+11)!9999999=1⇒(x+11)!=1⇒x+11=0or1⇒x=−11or−10⇒x1=−11,x2=−10orx1=−10,x2=−11⇒x1x2+0.9=1111+910=2or⇒x1x2+0.9=1011+910=199110
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