题目

题型：填空题难度：一般来源：不详

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

．