已知a+x2=2000,b+x2=2001,c+x2=2002,且abc=24,求a/bc+c/ab+b/ac−1/a−1/b−1/c的值.
问题描述:
已知a+x2=2000,b+x2=2001,c+x2=2002,且abc=24,求
+a bc
+c ab
−b ac
−1 a
−1 b
的值. 1 c
答
∵a+x2=2000,b+x2=2001,c+x2=2002,
∴b-a=1,c-b=1,c-a=2,
∵abc=24,
∴
+a bc
+c ab
−b ac
−1 a
−1 b
1 c
=
a2+ b2+c2−bc−ac−ab abc
=
2a2+2b2+2c2−2bc−2ac−2ab 2abc
=
(a−b)2+(a−c)2+(b−c)2
2abc
=
1+4+1 2×24
=
.1 8