Question

数学题！快！急用！

已知2005x3=2006y3=2007z3,xyz>0,且三次根号2005x2+2006y2+2007z2=三次根号2005+三次根号2006+三次根号2007,求1/x+1/y+1/z的值

Answer 1

设2005x^3=2006y^3=2007z^3=k
(2005x2+2006y2+2007z2)^(1/3)=2005^(1/3)+2006^(1/3)+2007^(1/3)
==>
(k/x+k/y+k/z)^(1/3)=(k/x^3)^(1/3)+(k/y^3)^(1/3)+(k/z^3)^(1/3)

==>k^(1/3)(1/x+1/y+1/z)^(1/3)=k^(1/3)(1/x+1/y+1/z)
==>1/x+1/y+1/z=(1/x+1/y+1/z)^(1/3)
==>1/x+1/y+1/z=1或0(舍)或-1(舍)
故1/x+1/y+1/z=1