已知a+x=2003,b+x=2004,c+x=2005,且abc=6012,求a/bc+b/ca+c/ab-1/a-1/b-1/c的值

a=2003-x^2,b=2004-x^2,c=2005-x^2 所以a=b-1,c=b+1 所以a/bc+b/ca+c/ab-1/a-1/b-1/c =a^2/(abc)+b^2/(abc)+c^2/(abc)-bc/(abc)-ac/(abc)-ca/(abc) =(a^2+b^2+c^2-bc-ac-ca)/abc =1/2[(a-b)^2+(a-c)^2+(b-c)^2]/abc =1/2*[1+4+1]/(abc) =3/6012 =1/2004