2014中考之分式的运算12答案

2018-12-17 11:53

班级 _______________________ 姓名_____________ 考场号__________ 考号_________

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一、选择题

1. D

2. A 3. A

二、填空题

4.

cb 5.

aa?1 6. 3 7. 0

三、计算题

28. 解:原式=?x?2??x?2??x?2??x?2x?2?x

=1?x

当x?1时原式=1?1?0

a29. 解:

?aba?b??a?b?0?b?tan60 ?a(a?b)a?b?1?b?3

?a?b?3 a?1,b?3,?原式?1?3?3??2

10. 解法一:原式=

(a?3)?(a?3)a(a?3)(a?3)?a2?9

=

2a(a?3)(a?(a?3)(a?3)?3)a =

2aa =2

班级 _______________________ 姓名_____________ 考场号__________ 考号_________

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a2?911解法二:原式=()? ?a?3a?3a

= = =

1(a?3)(a?3)1(a?3)(a?3)+ ??a?3aa?3aa?3a?3 ?aa2a a=2

a?1a?3(a?3)211. 解:原式= ??a?3a?2(a?2)(a?2)=

a?1a?3(a?2)(a?2)?· 2a?3a?2(a?3)a?1a?2? a?3a?33= a?3a取值时只要不取2,?2,3就可以. ?求值正确.

a?1a2?2a?1?12. 解:原式= aa =

a?1a ?2a?a?1?1 a?1 =

a取0和1以外的任何数,计算正确都可给分.

2xyx2?y2(x?y)2x?y13. 选择一:M?N?2, ???x?y2x2?y2(x?y)(x?y)x?y5y?y572当x∶y=5∶2时,x?y,原式=?. 52y?y322xyx2?y2?(x?y)2y?x选择二:M?N?2, ???x?y2x2?y2(x?y)(x?y)x?y 2

班级 _______________________ 姓名_____________ 考场号__________ 考号_________

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5y52??3. 当x∶y=5∶2时,x?y,原式=527y?y2y?x2?y22xy(x?y)2x?y选择三:N?M?2, ???x?y2x2?y2(x?y)(x?y)x?y5y?y53当x∶y=5∶2时,x?y,原式=2?.

52y?y72注:只写一种即可.

xx2?1?14. 解:原式= x?13xx(x?1)(x?1)? x?13xx?1=

3=

当x?1?3时,原式=

31?3?1= 33(x?1)2x215. 原式= ?(x?1)(x?1)x?1x?1x2?= x?1x?1x2?x?1=

x?1x2?2?0,

?x2?2

?原式?

2?x?1 x?1?原式=1

班级 _______________________ 姓名_____________ 考场号__________ 考号_________

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16. 解:原式?111x?y?(x?y)(x?y)?? 2xx?yx?y2x ?12x?(x?y)?12x ??(x?y) ?y?x

把x?2,y?3代入上式,得原式=3?2

17. 原式=(a?b)(a?b)a(a?b)???a2?2ab?b2??a??

=

a?ba·a(a?b)2 =1a?b

18. 解:原式=

(x?y)(x?y)(x?y)?2x?y

=x+y-2x+y =-x+2y 因为 x=3,y=2

所以原式=-3+4=1 .

19. 解:原式=1?(a?b)(a?b)a(a?b)?a

=1?a?b. 当a = 2,b??1时, 原式 = 2.

20. 解:原式=

(x?y)(x?y)x?y?2(x?y)

=x?y?2x?2y =?x?3y

当x?3,y??13时

班级 _______________________ 姓名_____________ 考场号__________ 考号_________

--------------------------------------------密--------------------封-------------------线---------------------------------------- 原式=?3?3?(?13)

=?2.

21. 解: ??xx?1?x2?1?x?1?x2?1???x2?x =??x?x?1?x?1?x2?1x2?1???x(x?1) =x2?1x(x?1)(x?1)(x?1)·x2?1 =

xx?1 当x??2009时,原式=?2009?2009?1?20092010.

22. 解:??1??1?x?1???1x2?1??x?2? =

xx?1·?x?1??x?1?1?x?2 ?x2?2

当x?2时,原式??2?2?2

?4

23. 解:

x2?2xy?y2x??2?xy?xy??y?x?? (x?y)2x2?y2?x(x?y)?xy ?x?yx·xy(x?y)(x?y) ?yx?y


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