Question

已知2005x^3=2006y^3=2007z^3,且1/x+1/y+1/z=1,说明3√2005x^2+2006y^2

Answer 1

设2005x^3=2006y^3=2007z^3=S^3

则2005=s^3/x^3,2006=s^3/y^3,2007=s^3/z^3.

左边=3√2005x^2+2006y^2+2007z^2=三次根号[(1/x+1/y+1/z)s^3]=s

右边=s/x+s/y+s/z=s(1/x+1/y+1/z)=s

所以原式成立.