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Question Number 66396 by hmamarques1994@gmai.com last updated on 14/Aug/19

    Seja  53^(log_(1/(√e^𝛑 )) [(((x+11)!))^(1/(9999999)) ])  = 1.            Calcule  (x_1 /x_2 )+0,9.

Seja53log1eπ[(x+11)!9999999]=1.Calculex1x2+0,9.

Answered by mr W last updated on 14/Aug/19

53^(log_(1/(√e^𝛑 )) [(((x+11)!))^(1/(9999999)) ])  = 1  ⇒log_(1/(√e^𝛑 )) [(((x+11)!))^(1/(9999999)) ]=0  ⇒(((x+11)!))^(1/(9999999)) =1  ⇒(x+11)!=1  ⇒x+11=0 or 1  ⇒x=−11 or −10  ⇒x_1 =−11, x_2 =−10 or x_1 =−10, x_2 =−11  ⇒(x_1 /x_2 )+0.9=((11)/(11))+(9/(10))=2  or  ⇒(x_1 /x_2 )+0.9=((10)/(11))+(9/(10))=((199)/(110))

53log1eπ[(x+11)!9999999]=1log1eπ[(x+11)!9999999]=0(x+11)!9999999=1(x+11)!=1x+11=0or1x=11or10x1=11,x2=10orx1=10,x2=11x1x2+0.9=1111+910=2orx1x2+0.9=1011+910=199110

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