已知：a+x²=1999,b+x²=2000,c+x²=2001,且abc=24,求a/b

a+1=b
b+1=c
a+2=c
a/bc+b/ac+c/ab-1/a-1/b-1/c
=(a²+b²+c²-bc-ac-ab)/abc
=[(a²-ac)+(b²-ab)+(c²-bc)]/abc
=[a(a-c)+b(b-a)+c(c-b)]/abc
=(-2a+b+c)/abc
=[(b-a)+(c-a)]/abc
=(1+2)/abc
=3/abc
=3/24
=1/8