题目内容：

数学题：难题
设2006x3=2007y3=2008z3,且xyz＞0,3√2006x2+2007y2+2008z2 =3√2006+3√2007+ 3√2008,求1／x+1／y+1／z

优质解答

设2006x^3=2007y^3=2008z^3=k(2006x2+2007y2+2008z2)^(1/3)=2006^(1/3)+2007^(1/3)+2008^(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/...