若a+x^2=1998,b+x^2=1999,c+x^2=2000,abc=24 求c/ab+b/ac+a/bc-1/a-1/b-1/c的值

题目

若a+x^2=1998,b+x^2=1999,c+x^2=2000,abc=24 求c/ab+b/ac+a/bc-1/a-1/b-1/c的值

答案

由a+x^2=1998,b+x^2=1999,c+x^2=2000得:
a=1998-x^2
b=1999-x^2
c=2000-x^2
即:a,b,c是连续的3个自然数,又abc=24
所以:a=2
b=3
c=4
a/bc+c/ab+b/ac-1/a-1/b-1/c
=a^2/abc+c^2/abc+b^2/abc-1/a-1/b-1/c
=4/24+16/24+9/24-1/2-1/3-1/4
=1/6+2/3+3/8-1/2-1/3-1/4
=1/8