若x^2+y^2+z^2=2x+6y+8z-26,求xyz/x^2+y^2+z^2的值

题目

若x^2+y^2+z^2=2x+6y+8z-26,求xyz/x^2+y^2+z^2的值

答案

x^2+y^2+z^2=2x+6y+8z-26
x^2+y^2+z^2-2x-6y-8z+26=0
(x^2-2x+1)+(y^2-6y+9)+(z^2-8z+16)=0
(x-1)^2+(y-3)^2+(z-4)^4=0
所以x-1=0,y-3=0,z-4=0
所以x=1,y=3,z=4
所以xyz/(x^2+y^2+z^2)
=1*3*4/(1^2+3^2+4^2)
=12/26
=6/13