Question

已知a，b，c是非零有理數(shù)，且滿足ab2=

c

a

-b，則(

a2b2

c2

-

2

c

+

1

a2b2

+

2ab

c2

-

2

abc

)÷(

2

ab

-

2ab

c

)÷

101

c

等于

 

．

Answer 1

分析：先在等式ab2=

c

a

-b兩邊同乘非零數(shù)a，得到a²b²=c-ab，移項(xiàng)得出a²b²-c=-ab，c-a²b²=ab．再將得到的三個(gè)等式代入所求代數(shù)式，然后化簡，即可得出結(jié)果．

解答：解：∵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

．
故答案為-

1

202

．

點(diǎn)評(píng)：本題考查了分式的化簡求值，屬于競(jìng)賽題型，難度較大．將已知等式變形是關(guān)鍵，將所求代數(shù)式分項(xiàng)組合使之能夠應(yīng)用已知條件是難點(diǎn)．