2019-2020年九年级数学综合练习(一)
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2019-2020年九年级数学综合练习（一）
一、选择题（本大题共10小题，每小题3分）
11. 1； 12.4; 13.2
(2)a x -；14.相交；15.<；16.50.
三、解答题（本大题共9小题，满分102分．解答应写出文字说明、证明过程或演算步骤） 三、17．解：（1）B 所对应的实数为4, ······························································ 2分 8.AB = ·
······································································································ 4分 （2）由题意得，22
431
x x +=-，·
········································································· 6分 解得3
5
x =． ································································································· 8分
经检验，3
5x =是原方程的解．
∴x 的值为3
5
． ····························································································· 9分
三、18.解：5125431x x x x ->+⎧⎨-<+⎩
，①
②
由①得2x >， ······························································································· 3分 由②得，52
x >-
···························································································· 5分 ∴原不等式组的解集为2x > ··········································································· 7分
不等式的解集在数轴上表示如图. ····································································· 9分
三、19证明（1）
DE AB DF AC ⊥，⊥，
90BED CFD ∴∠=∠=°， ············································································· 1分 AB AC =，B C ∴∠=∠， ··········································································· 3分 D 是BC 的中点，
BD CD ∴=， ·
······························································································ 4分 BED CFD ∴△≌△. ······················································································· 5分 （2）四边形DFAE 为正方形． ········································································· 7分
DE AB DF AC ⊥，⊥， 90AED AFD ∴∠=∠=°，
90A ∠=°，∴四边形DFAE 为矩形. ·
······························································ 9分 又BED CFD △≌△，
DE DF ∴=，∴四边形DFAE 为正方形． ·
······················································ 10分
三、20．解（1）法一：根据题意，可以画出如下的树形图：
········································· 5分 图12
D
C E
A
F B 1 2 3
2 1
3 3 1 2 第一个球
第二个球
从树形图可以看出，摸出两球出现的所有可能结果共有6种； 法二：根据题意，可以列出下表：
从上表中可以看出，摸出两球出现的所有可能结果共有6种. ···························· 5分 （2）设两个球号码之和是奇数为事件A .
摸出的两个球号码之和是奇数的结果有4种，它们是：()23，、(1,2)、(3,2)、(2,1)
()42
63
P A ∴=
= ·
············································································· 10分 三、21．解：设生产亚运会标志x 套，生产亚运会吉祥物y 套． ····························· 2分
根据题意，得0.40.52300............0.33600..............x y ,x y .
+=⎧⎨
+=⎩①
② ·························· 6分
①×2－②×1得：0.5x =1000．
∴ x =2000． ······················································································· 9分 把x =2000代入②得：600+y =3600． ∴ y =3000． ······················································································· 11分
答：该厂能生产亚运会标志2000套，生产亚运会吉祥物3000套． ··················· 12分 三、22．解：（1）直线BD 与⊙O 相切．… 1分 证明：如图1，连结OD ． …………… 2分
OA OD =，A ADO ∴∠=∠．…………… 3分 90C ∠=， 90CBD CDB ∴∠+∠=．
又
CBD A ∠=∠， ………………………… 5分 90ADO CDB ∴∠+∠=．
180()90ODB ADO CDB ∴∠=-∠+∠=．∴直线BD 与⊙O 相切． ···················· 6分
（2）连OD 、DE
AD BD = A DBA ∴∠=∠． ·············································· 7分 在Rt BDC ∆中90C ∠=，CBD A DBA ∠=∠=∠， 390A ∴∠=,即有=30A ∠ ·············································································· 8分 由tan DE
A AD
∠=
，得tan 302DE AD =⋅=⨯=． ································· 10分 又60,DOE OD OE ∠==,DOE ∴∆为等边三角形
,OD DE ∴==
. ·············· 10分 即⊙O
的半径r OD ==
，故⊙O 的面积2
4.3
S r ππ== ································ 12分
三、23．解:（1）若1m =, 方程化为2
540x x -+= ············································· 2分
第二个球 第一个球
（1，3） （2，3）
（1，2）
（3，2） （3，1）
（2，1） 3 2 1
1
2
3