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2010年中考数学考前50天得分专练24

一、填空题（本大题每个空格1分，共18分．把答案填在题中横线上）

2的相反数是 1．

2．点A1(，2 )

1

的绝对值是，立方等于 64的数是．3

关于x轴对称的点的坐标是；点A关于原点对称的点的坐标是．

3．若∠ 30，则∠ 的余角是 °，cos ．

4．在校园歌手大赛中，七位评委对某位歌手的打分如下：9.8，9.5，9.7，9.6，9.5，9.5，9.6，

则这组数据的平均数是 ，极差是 ．

5．已知扇形的半径为2cm，面积是为 °．

4

cm2，则扇形的弧长是，扇形的圆心角3

k ． 2)，B(1，0)，6．已知一次函数y kx b的图象经过点A(0，则b ，

7．如图，已知DE∥BC，AD 5，DB 3，BC 9.9，

∠B 则∠ADE °，DE ，

S△ADE

S△ABC

B（第7题）

8．二次函数y ax2 bx c的部分对应值如下表：

二次函数y ax2 bx c图象的对称轴为x x 2对应的函数值

y 二、选择题（下列各题都给出代号为A，B，C，D的四个答案，其中有且只有一个是正确的，把正确答案的代号填在题后（ ）内，每小题2分，共18分）9．在下列实数中，无理数是（ ）

A．

1 3

B．

CD．

227

10．在函数y A．x 2

1

中，自变量x的取值范围是（ ）x 2

B．x≤ 2

C．x 2

D．x≥ 2

11．下列轴对称图形中，对称轴的条数最少的图形是（ ）A．圆 B．正六边形 C．正方形 D．等边三角形

12．袋中有3个红球，2个白球，若从袋中任意摸出1个球，则摸出白球的概率是（ ）

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A．

1 5

B．

2 5

C．

2 3

D．

13

13．如图，图象（折线OEFPMN）描述了某汽车在行

驶过程中速度与时间的函数关系，下列说法中错误 ．．

的是（ ）

A．第3分时汽车的速度是40千米/时

B．第12分时汽车的速度是0千米/时

C．从第3分到第6分，汽车行驶了120千米D．从第9分到第12分，汽车的速度从60千米/时

减少到0千米/时

14．下面各个图形是由6个大小相同的正方形组成的，其中

能沿正方形的边折叠成一个正方体的是（ ）

/分 （第13题） A． B． C． D．

15．小明和小莉出生于1998年12月份，他们的出生日不是同一天，但都是星期五，且小明

比小莉出生早，两人出生日期之和是22，那么小莉的出生日期是（ ） A．15号 B．16号 C．17号 D．18号 16．若二次函数y ax bx a 2（a，b为常数）的图象如下，则a的值为（ ）

A． 2

B

．

C．1

D

2

2

CA

（第16题）

（第17题）

17．如图，在△ABC中，AB 10，AC 8，BC 6，经过点C且与边AB相切的动圆

与CA，CB分别相交于点P，Q，则线段PQ长度的最小值是（ ） A．4.75

B．4.8

C．5

D

．三、解答题（本大题共2小题，共18分．解答应写出演算步骤） 18．（本小题满分10分）化简： （1

）2 2 （2）

2

41

． x2 4x 2

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19．（本小题满分8分）解方程： （1）

34

； （2）x2 2x 2 0． x 1x

20．（本小题满分5分）

已知，如图，在ABCD中，∠BAD的平分线交BC边于点E． 求证：BE CD．

B

（第20题）

21．（本小题满分7分）

已知，如图，延长△ABC的各边，使得BF AC，AE CD AB，顺次连接D，E，F，得到△DEF为等边三角形．

求证：（1）△AEF≌△CDE；

（2）△ABC为等边三角形．

E（第21题）

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参考答案

一、填空题（每个空格1分，共18分）

1．2，5．

1

， 4； 3

2．(12)； ，2)，( 1，3．60

4．9.6，0.3；

4

，120； 3

6． 2，2； 7．50，6.6，

4

； 9

8．1， 8．

三、解答题（本大题共2题，第18题10分，第19题8分，共18分．解答应写出演算步骤）

18．解：（1）原式 1 （2）原式

1

3 ································································································ 3分 4

7． ···································································································· 5分 4

4x 2

········································································· 2分

(x 2)(x 2)(x 2)(x 2)

4 x 2

···································································································· 3分

(x 2)(x 2)

(x 2)

···································································································· 4分

(x 2)(x 2)

1

． ··········································································································· 5分 x 2

19．解：（1）去分母，得3x 4x 4． ··············································································· 1分 解得，x 4． ························································································································ 2分 经检验，x 4是原方程的根．

······································································································· 4分 原方程的根是x 4． ·

（2）(x 1) 3， ··

·············································································································· 2分 ·········

··············

································································································ 3分 x 1 ·

························································································ 4分 x1 1x2 1 ·

21．证明： 四边形ABCD是平行四边形， AD∥BC，AB CD．

∠DAE ∠BEA． ············································································································ 1分

2

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AE平分∠BAD， ∠BAE ∠DAE． ······································································· 2分 ∠BAE ∠BEA． ············································································································ 3分 AB BE． ························································································································ 4分 又 AB CD， BE CD．···························································································· 5分 21．证明：（1） BF AC，AB AE， FA EC． ··············································· 1分 △DEF是等边三角形， EF DE． ············································································ 2分 又 AE CD， △AEF≌△CDE． ·············································································· 4分 （2）由△AEF≌△CDE，得∠FEA ∠EDC，

∠BCA ∠EDC ∠DEC ∠FEA ∠DEC ∠DEF，△DEF是等边三角形，

∠DEF 60 ，

∠BCA 60 ，同理可得∠BAC 60 ．········································································· 5分 △ABC中，AB BC． ··································································································· 6分 △ABC是等边三角形． ······································································································ 7分