Question

已知a+x²=2003,b+x²=2004,c+x²=2005,且abc=6012,求a/bc+b/ca+c/ab-1/a-1/b-1/c的值

Answer 1

易知:b-a=1, c-b=1, c-a=2 .
a/bc+b/ca+c/ab-1/a-1/b-1/c
=(a/bc-1/a)+(b/ca-1/b)+(c/ab-1/c)
=(a^2-bc)/abc+(b^2-ac)/abc+(c^2-ab)/abc
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)］/2*abc
=(1+1+4)/2*6012
=1/2004

Answer 2

∵a+x²=2003,b+x²=2004,c+x²=2005
∴a-b=-1 b-c=-1 c-a=2
a/bc+b/ac+c/ab-1/a-1/b-1/c
=(a/bc-1/a)+(1/ac-1/b)+1/ab-1/c)
=(a²-bc)/abc+(b²-ac)/abc+(c²-ab)/abc
=(a²+b²+c²-ab-ac-bc)/abc
=[(a-b)²+(b-c)²+(c-a)²]/6012
=[(-1)²+(-1)²+2²]/6012
=4/6012
=1503