问题 填空题
已知a,b,c是非零有理数,且满足ab2=
c
a
-b
,则(
a2b2
c2
-
2
c
+
1
a2b2
+
2ab
c2
-
2
abc
)÷(
2
ab
-
2ab
c
101
c
等于______.
答案

ab2=

c
a
-b,

∴a2b2=c-ab,a2b2-c=-ab,c-a2b2=ab.

a2b2
c2
-
2
c
+
1
a2b2
+
2ab
c2
-
2
abc
=(
ab
c
-
1
ab
2+
2a2b2
abc2
-
2c
abc2
=(
a2b2-c
abc
2+
2(a2b2-c)
abc2
=(
-ab
abc
2+
-2ab
abc2
=
1
c2
-
2
c2
=-
1
c2

2
ab
-
2ab
c
=
2c-2a2b2
abc
=
2ab
abc
=
2
c

(

a2b2
c2
-
2
c
+
1
a2b2
+
2ab
c2
-
2
abc
)÷(
2
ab
-
2ab
c
101
c
=-
1
c2
÷
2
c
÷
101
c
=-
1
c2
c
2
c
101
=-
1
202

故答案为-

1
202

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单项选择题