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Question-120464




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
$$\sqrt{{x}−\mathrm{1}}={a}\:\:\Rightarrow\:{x}={a}^{\mathrm{2}} +\mathrm{1} \\ $$$$\sqrt{{y}−\mathrm{4}}={b}\:\Rightarrow\:{y}={b}^{\mathrm{2}} +\mathrm{4} \\ $$$$\sqrt{{z}−\mathrm{9}}={c}\:\Rightarrow\:{z}={c}^{\mathrm{2}} +\mathrm{9}\: \\ $$$${x}+{y}+{z}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{14} \\ $$$${a}+\mathrm{2}{b}+\mathrm{3}{c}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{14}}{\mathrm{2}}\:\:\mid×\mathrm{2} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{14}=\mathrm{2}{a}+\mathrm{4}{b}+\mathrm{6}{c} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}+{b}^{\mathrm{2}} −\mathrm{4}{b}+\mathrm{4}+{c}^{\mathrm{2}} −\mathrm{6}{c}+\mathrm{9}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\left({b}−\mathrm{2}\right)^{\mathrm{2}} +\left({c}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\mathrm{1}\:{b}=\mathrm{2}\:{c}=\mathrm{3}\:\:\Rightarrow\:{x}+{y}+{z}=\mathrm{28} \\ $$

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