Question

两道数学题，在线等，希望大哥大姐们帮帮忙啊

1.已知a+x²=2007，b+x²=2008，c+x²=2009，abc=6012
求a/bc+b/ca+c/ab-1/a-1/b-1/c
2.设abc=1，求分式a/ab+a+1+b/bc+b+1+c/ca+c+1的值
请各位写出详细过程啊……我们明天要交啊！

Answer 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)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

Answer 2

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)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

Answer 3

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？

Answer 4

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