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If-a-1-b-7-3-b-1-c-4-c-1-a-1-find-2022-abc-

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Question Number 184431 by SulaymonNorboyev last updated on 06/Jan/23
If { ((a+(1/b)=(7/3))),((b+(1/c)=4)),((c+(1/a)=1)) :} find 2022∙abc
If{a+1b=73b+1c=4c+1a=1find2022⋅abc
Answered by mr W last updated on 06/Jan/23
c=1−(1/a)=((a−1)/a) b=4−(a/(a−1))=((3a−4)/(a−1)) ⇒abc=a×((3a−4)/(a−1))×((a−1)/a)=3a−4 a+((a−1)/(3a−4))=(7/3) 9a^2 −30a+25=0 (3a−5)^2 =0 ⇒a=(5/3) abc=3a−4=5−4=1 2023abc=2023 ✓
c=1−1a=a−1ab=4−aa−1=3a−4a−1⇒abc=a×3a−4a−1×a−1a=3a−4a+a−13a−4=739a2−30a+25=0(3a−5)2=0⇒a=53abc=3a−4=5−4=12023abc=2023✓
Answered by ajfour last updated on 06/Jan/23
(a+(1/b))(b+(1/c))(c+(1/a))=pqr abc+b+a+(1/c)+c+(1/a)+(1/b)+(1/(abc)) =pqr abc+(1/(abc))+(p+q+r)=pqr (abc)((1/(abc)))=1 ⇒ abc, (1/(abc))=((pqr−(p+q+r))/2) ±(√({((pqr−(p+q+r))/2)}^2 −1)) =((14)/3)−((11)/3)±(√(1−1)) =1
(a+1b)(b+1c)(c+1a)=pqrabc+b+a+1c+c+1a+1b+1abc=pqrabc+1abc+(p+q+r)=pqr(abc)(1abc)=1⇒abc,1abc=pqr−(p+q+r)2±{pqr−(p+q+r)2}2−1=143−113±1−1=1
Commented by mr W last updated on 06/Jan/23
very smart approach!
verysmartapproach!
Commented by manxsol last updated on 07/Jan/23
(abc)((1/(abc)))=1 to the toolbox
(abc)(1abc)=1tothetoolbox
Commented by manxsol last updated on 07/Jan/23
a surreal answer
asurrealanswer

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