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Question Number 120166 by Algoritm last updated on 29/Oct/20

Commented by Algoritm last updated on 29/Oct/20

what ? ):

what?):

Answered by nimnim last updated on 30/Oct/20

a+b+c+d=S..................(1) (1/a)+(1/b)+(1/c)+(1/d)=S..........(2) (1/(a−1))+(1/(b−1))+(1/(c−1))+(1/(d−1))=S......(3) a+(1/a)+(1/(a−1))=((a(a−1)+a−1+1)/(a(a−1)))=(a^2 /(a(a−1)))=(a/(a−1)) ∴ (a/(a−1))+(b/(b−1))+(c/(c−1))+(d/(d−1))=3S ⇒((a/(a−1))−1)+((b/(b−1))−1)+((c/(c−1))−1)+((d/(d−1))−1)=3S−4 ⇒(1/(a−1))+(1/(b−1))+(1/(c−1))+(1/(d−1))=3S−4......(4) From (3) and (4) 3S−4=S ⇒S=2★

a+b+c+d=S..................(1)1a+1b+1c+1d=S..........(2)1a−1+1b−1+1c−1+1d−1=S......(3)a+1a+1a−1=a(a−1)+a−1+1a(a−1)=a2a(a−1)=aa−1∴aa−1+bb−1+cc−1+dd−1=3S⇒(aa−1−1)+(bb−1−1)+(cc−1−1)+(dd−1−1)=3S−4⇒1a−1+1b−1+1c−1+1d−1=3S−4......(4)From(3)and(4)3S−4=S⇒S=2★

Commented by Algoritm last updated on 30/Oct/20

S=−2

S=−2

Commented by Algoritm last updated on 30/Oct/20

Commented by nimnim last updated on 30/Oct/20

okay...

okay...

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