1.已知a+x²=2007,b+x²=2008,c+x²=2009,abc=6012

原式=(a/bc-1/a)+(b/ca-1/b)+(c/ab-1/c)
=(a²-bc)/abc+(b²-ac)/abc+(c²-ab)/abc
=(a²+b²+c²-bc-ac-ab)/abc
=[1/2(a-b)²+1/2(b-c)²+1/2(a-c)²]/abc
a+x²=2007,b+x²=2008,c+x²=2009
a+x²-b-x²=2007-2008
a-b=-1
a+x²-c-x²=2007-2009
a-c=-2
b+x²-c-x²=2008-2009
b-c=-1
abc=6012
原式=[1/2(-1)²+1/2(-1)²+1/2(-2)²]/6012
=3/6012
=1/2004

1.原式=(a/bc-1/a)+(b/ca-1/b)+(c/ab-1/c)
=(a²-bc)/abc+(b²-ac)/abc+(c²-ab)/abc
=(a²+b²+c²-bc-ac-ab)/abc
=[1/2(a-b)²+1/2(b-c)&su...

a+x²=2007，b+x²=2008，c+x²=2009 得出 a+2=b+1=c
a/bc+b/ca+c/ab-1/a-1/b-1/c
=(a²+b²+c²-ab-ac-bc)/abc 把a=c-2,b=c-1带入化简得
原式=3/abc=1/2004
第二道是不是少括号了？

a/bc+b/ca+c/ab-1/a-1/b-1/c
=（(a-c)2+(b-c)2+(a-b)2）/2abc
=1/2004