若a+x2=2007,b+x2=2008,c+x2=2009,且abc=24,则abc+bac+cab−1a−1b−1c
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若a+x2=2007,b+x2=2008,c+x2=2009,且abc=24,则
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![若a+x2=2007,b+x2=2008,c+x2=2009,且abc=24,则abc+bac+cab−1a−1b−1c](/uploads/image/z/15081282-18-2.jpg?t=%E8%8B%A5a%2Bx2%3D2007%EF%BC%8Cb%2Bx2%3D2008%EF%BC%8Cc%2Bx2%3D2009%EF%BC%8C%E4%B8%94abc%3D24%EF%BC%8C%E5%88%99abc%2Bbac%2Bcab%E2%88%921a%E2%88%921b%E2%88%921c)
∵a+x2=2007,b+x2=2008,c+x2=2009,
∴b-a=1,c-b=1,c-a=2,
∴
a
bc+
b
ac+
c
ab−
1
a−
1
b−
1
c=
a
bc−
1
b+
b
ac−
1
c+
c
ab−
1
a
=
a−c
bc+
b−a
ac+
c−b
ab
=
−2
bc+
1
ac+
1
ab
=
−2a+b+c
abc
=
(b−a)+(c−a)
abc
=
1+2
24
=
1
8.
∴b-a=1,c-b=1,c-a=2,
∴
a
bc+
b
ac+
c
ab−
1
a−
1
b−
1
c=
a
bc−
1
b+
b
ac−
1
c+
c
ab−
1
a
=
a−c
bc+
b−a
ac+
c−b
ab
=
−2
bc+
1
ac+
1
ab
=
−2a+b+c
abc
=
(b−a)+(c−a)
abc
=
1+2
24
=
1
8.
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