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Question Number 120464 by pooooop last updated on 31/Oct/20

Answered by Jamshidbek2311 last updated on 31/Oct/20

(√(x−1))=a ⇒ x=a^2 +1 (√(y−4))=b ⇒ y=b^2 +4 (√(z−9))=c ⇒ z=c^2 +9 x+y+z=a^2 +b^2 +c^2 +14 a+2b+3c=((a^2 +b^2 +c^2 +14)/2) ∣×2 a^2 +b^2 +c^2 +14=2a+4b+6c a^2 −2a+1+b^2 −4b+4+c^2 −6c+9=0 (a−1)^2 +(b−2)^2 +(c−3)^2 =0 a=1 b=2 c=3 ⇒ x+y+z=28

x−1=a⇒x=a2+1y−4=b⇒y=b2+4z−9=c⇒z=c2+9x+y+z=a2+b2+c2+14a+2b+3c=a2+b2+c2+142∣×2a2+b2+c2+14=2a+4b+6ca2−2a+1+b2−4b+4+c2−6c+9=0(a−1)2+(b−2)2+(c−3)2=0a=1b=2c=3⇒x+y+z=28

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