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5.在一个不透明的袋子中装有两个红球，一个白球，一个黄球，这些球除颜色外都相
同.从中一次摸出两个球，记下颜色后放回，多次重复上述试验，摸到的两个球颜色
相同的频率最可能接近的数值为
A.1.2
B.0.78
C.0.52
D.0.17
数
学
6.下面解分式方程1-x=
2
x-3=3-
一的步骤中，错误的是
A.将方程两边同时乘x-3可转化为整式方程
注意事项：
B.去分母后的一元一次方程为1-x=-2
1.本试卷共8页，满分120分，考试时间120分钟.
C.原分式方程的解为x=3
2.答卷前，考生务必将自己的姓名、准考证号填写在本试卷相应的位置
D.原分式方程无解
3.答案全部在答题卡上完成，答在本试卷上无效
7.开窗见蓝天、出门迎白云，如今已成为太原市民的日常.一组数据印证了这份“蓝天
4.考试结束后，将本试卷和答题卡一并交回
幸福感”：2023年全年太原市空气污染天数为121天，2025年全年太原市空气污染
天数为82天.设连续两年太原市空气污染天数的平均减少率为x,则下面所列方程
第I卷选择题（共30分）
正确的是
A.121(1-x)2=82
B.121(1-x%)2=82
一、选择题（本大题共10个小题，每小题3分，共30分.在每个小题给出的四个选项中，
C.82(1+x)2=121
D.121(1-2x)=82
只有一项符合题目要求，请选出并在答题卡上将该项涂黑)
1.二十四节气是中国古代农耕文明的重要组成部分，用来指导农业生产和日常生活
8.如图，E,F是口ABCD中DA,BC延长线上的两点，连接EF分别
乐乐查询了当地2025年大寒时的最高气温为-7℃，大暑时的最高气温为30℃，则
交AB,CD于点G,H.下列各式中正确的是
两个节气的最高气温相差
AC剧
B.
EG BC
GH AE
A.37℃
B.33℃
C.30℃
D.23℃
DH AG
第8题图
2.山西省，简称“晋”，又称“三晋”.下面是小明收集的关于“晋”字的演变过程，其文字
c治-所
D.
CH=BG
上方的部分是轴对称图形的是
9.已知点A,B是一次函数y=-mx+n(m≠0)图象上的两点，若y,亚
确的是
A.xB.>x
C.x1=x习
D.无法判断
甲骨文
金文
小篆
楷书
10.如图，AB是⊙0的直径，AD,BC,AC是⊙0的弦.若∠BAD=30°，
A
B
D
BD=BC,AC=3√3，则AC长为
3.下列运算正确的是
3
A.xx=x
B.(x-1)(x-1)=x2+1
A.2T
B.2T
C.(-m2)3=-m6
D.m2+m'=m
C.√3m
D.3√3T
第10题图
4.位于朔州市的应县木塔，是“世界三大奇塔”之一，也
第Ⅱ卷非选择题（共90分）
是世界上现存最古老的纯木结构楼阁式建筑，它的结
构独特，全榫卯连接，没有一根铁钉.右图是一种简单
二、填空题（本大题共5个小题，每小题3分，共15分）
A
的榫卯结构示意图，几何体A和B组合后得到新的几
正面
11.比较大小：√5▲2（填“>”“=”或“<”）.
何体，其中几何体A的俯视图为
第4题图
12.如图，△ABC的顶点B,C的坐标分别为(1，-1)，(2,1)，将
△ABC沿CB方向平移得到△DEB,若点A的对应点D的坐
B
标是(-2,0)，则点A的坐标为▲
第12题图
数学试卷（二）第1页（共8页）
数学试卷（二）第2页（共8页）数学参考答案与评分标准
一、选择题
1 2 3 4 5 6 7 8 9 10
A C C B D C A A B B
二、填空题
11. ＞ 12. （-1，2） 13. 9 14. 100
15. 2
提示：如图，连接 BG，过点 G 作 GM⊥BC 于点 M，过点 G 作 GN⊥DC，交
DC的延长线于点 N.
∵EF垂直平分 BD，∴BG=DG.∴∠DBG=∠BDG=45°.
∴∠GDN+∠ADB=90°-∠BDG=45°.
由矩形易得∠ADB=∠DBC，
∵∠GBM=45°-∠DBC，∠GDN=45°-∠ADB，
∴∠GDN=∠GBM.
∵BG=DG，∠BMG=∠DNG=90°，
∴△BMG≌△DNG.∴GM=GN，BM=DN.
易证四边形MGNC是正方形，设正方形边长为 x，
则 BM=4-x，DN=2+x.∴4-x=2+x.解得 x=1.∴CG= 2 .
三、解答题
16.解：
（1）原式=-3-9+（-2）································································································· 3分
=-14.················································································································ 5分
2 = a 1 a 1
2
（ ）原式 ·······························································································7分
a a a 1
a 1 a a 1= 2 ·······························································································8分a a 1
=1.···················································································································· 10分
17.证明：∵CE⊥AD，DF⊥BC，
∴∠CED=90°，∠F=90°.·························································································1分
∵四边形 ABCD是平行四边形，
∴AD∥BC.··············································································································· 2分
∴∠ECF+∠CED=180°.
∴∠ECF=90°.··········································································································· 3分
∴四边形 CEDF是矩形.·························································································· 4分
∵∠ADC=45°，
∴∠ECD=90°-45°=45°.··························································································· 5分
∴∠ADC=∠ECD.
∴DE=CE.··················································································································6分
∴矩形 CEDF是正方形.·························································································· 7分
18.解：（1）设单目显微镜的单价为 x元/台，双目显微镜的单价为 y元/台.··········· 1分
3x 2y=1440，
根据题意，得 ··················································································3分
8x 5y=3720.
x=240，
解，得 ·········································································································4分
y=360.
答：单目显微镜的单价为 240元/台，双目显微镜的单价为 360元/台.················ 5分
（2）设可购买双目显微镜 m台.
根据题意，得 240（30-m）+360m≤10 000.·····························································6分
解，得 m 70≤ .
3
因为 m为整数，且 m取最大值，所以 m=23.··························································7分
答：最多可购买双目显微镜 23台.·············································································8分
19.解：（1）①C················································································································1分
③D·····························································································································2分
④88.8·························································································································3分
（2）补充完整的统计图，如图所示：······································································5分
20 6
（3）600 =312（名）.
50
答：估计有 312名学生能完成目标.·······································································6分
目标合理.
理由：因为 40%+12%=52%，过半的学生都能完成这个目标，所以这个目标
合理.···························································································································7分
（4）①学校“阳光体育运动”采取的措施成果显著，超过一半的学生平均每天
体育运动的时间为 90分钟；②仍然有 18%学生每天平均体育运动的时间不
足一小时，学校还需提供更多的体育运动机会.（合理即可）···························9分
20.解：如图 1，延长 DE交 AB于点M；如图 2，延长 PQ交 AB于点 N.············· 1分
图 1 图 2
则四边形 DCBM，四边形 PCBN都是矩形.
∴BM=DC=1.5 m，BN=CP=10.5 m，DM=PN=BC.
设 AB=x m，则 AM=（x-1.5）m，AN=（x-10.5）m.········································2分
在 Rt△ADM中，∠AMD=90°，∠ADM=50°，
∴tan∠ADM= AM .
DM
DM= AM x 1.5∴ .··················································································· 3分
tan ADM 1.19
在 Rt△APN中，∠ANP=90°，∠APN=37°，
∴tan∠APN= AN ，
PN
PN= AN x 10.5∴ .·················································································· 5分
tan APN 0.75
x 1.5 x 10.5
由 DM=PN,得 .
1.19 0.75
解，得 x≈25.8.·········································································································· 6分
答：城门顶部点 A到地面的距离 AB的长约为 25.8米.····································· 7分
21.解：（1）∵∠BCD=∠AFE，
∴180°-∠BCD=180°-∠AFE.
∴∠3=∠4.················································································································ 1分
∵∠H=180°-∠3-∠2，∠G=180°-∠4-∠1，
∴∠H=∠G.·············································································································· 2分
∵∠ABC=∠DEF，
∴∠H+∠ABC=∠G+∠DEF.·················································································· 3分
∵∠H+∠ABC+∠G+∠DEF=360°，
∴∠H+∠ABC=180°.······························································································· 4分
∴BG∥HE，
即 AB∥DE.···············································································································5分
同理可得 BC∥EF，CD∥AF.
∴凸六边形 ABCDEF是平行六边形.·····································································6分
（2）是 ····················································································································7分
（3）如图，六边形 AEFCGH即为所求.
··················································································9分
22.解：（1）∵线段 AB和线段 CD互相垂直平分，且 AB=CD=6，
∴A（-3，3），B（3，3），C（0，6）.································································ 1分
由已知，点 C为抛物线的顶点，
∴设抛物线的函数表达式为 y=ax2+6.····································································2分
将 A（-3，3）代入，得 9a+6=3.
解，得 a= 1 .··········································································································· 3分
3
1
∴抛物线的函数表达式为 y x2 6 .································································ 4分
3
: 2 y=0 1（ ）当 时， x2 6 0.
3
解，得 x1 3 2,x 2 3 2. ·····················································································5分
∴ E( 3 2,0), F (3 2,0).
∴EF 6 2 cm.······································································································6分
由已知，得点 G和点 H的纵坐标为 4.
1
当 y=4时， x2 6 4.
3
解，得 x1 6,x 2 6. ························································································ 7分
∴G( 6,0),H ( 6,0).
∴GH 2 6 cm.·····································································································8分
3 2 4 17（ ）① n m2 m . ·················································································10分
3 3 3
8 5
②窗框的总长度为 (4 ) cm. ·················································································12分
5
23.解：（1）四边形 CDGF是菱形.·················································································1分
理由如下：由旋转得∠DCF=120°，∠F=∠BDC=60°，DC=FC，
∴∠DCF+∠F=180°.
∴DC∥GF.················································································································ 2分
∵∠DBC=90°，
∴∠BCD=90°-∠BDC=30°.
∵△ABC是等边三角形，
∴∠ACB=60°，AC=BC.
∴∠ACD=∠ACB - ∠BCD=30°.
∴∠ACD=∠BCD.
∵CD=CD，
∴△ACD≌△BCD.··································································································3分
∴∠ADC=∠BDC=60°.
∴∠ADC+∠DCF=180°.
∴DG∥CF.
∴四边形 CDGF是平行四边形.··············································································4分
∵DC=FC，
∴四边形 CDGF是菱形.··························································································5分
（2）AD=EG.理由如下：
如图 1，延长 DC到M，使 DM=DB，连接 BM，CG.
图 1
∵∠BDM=60°，
∴△DBM是等边三角形.···························································································· 6分
∴∠DBM=∠M=60°，BD=BM.
∵△ABC是等边三角形，
∴∠ACB=∠ABC=60°，AB=BC.
∴∠ABC - ∠DBC=∠DBM -∠DBC.
∴∠ABD=∠CBM.
∴△ABD≌△CBM.····································································································· 7分
∴∠ADB=∠M=60°.
∵∠BDC=60°，
∴∠CDG=180°-60°-60°=60°.
由旋转得∠DCF=120°，∠F=∠BDC=60°，CD=CF，
∴∠DCF+∠F=180°，∠DCF+∠CDG=180°.
∴CF∥DG，CD∥GF.
∴四边形 CDGF是平行四边形.··················································································8分
∵CD=CF，
∴四边形 CDGF是菱形.
∴DC=DG.
∴△DCG是等边三角形.····························································································· 9分
∴DC=GC，∠DCG=60°.
由旋转得∠BCE=120°，AC=EC，
∵∠ACB=60°，
∴∠ACE=60°.
∴∠ACE-∠DCE=∠DCG-∠DCE，即∠ACD=∠ECG.
∴△ACD≌△ECG.
∴AD=EG.····················································································································· 10分
（3 1 1） 或 .······················································································································13分
2
（答对 1个 2分）
提示：分∠DCH=90°和∠DHC=90°两种情况讨论.
情况一：当∠DCH=90°时，如图 2,∠ECF=∠DCF-∠DCH=30°.
∵四边形 CDGF是菱形，∴GF∥DC.∴∠FEC=∠DCH=90°.
∴EF= 1 FC. EF= 1 GF. EG=EF. EG∴ ∴ ∴ =1.
2 2 EF
图 2 图 3
情况二：当∠DHC=90°时，如图 3,∠DCH=90°-∠CDH=30°.
1
∴DH= DC.∴DH= 1 DG，DH= 1 CF.∴DH=HG，HG= 1 CF.
2 2 2 2
∵四边形 CDGF是菱形，∴DG∥CF.∴△EHG∽△ECF. EG = HG 1∴ .
EF CF 2
EG 1
综上所述， 的值为 1或 .
EF 2
【说明】以上各题的其他解法，请参照此标准评分.

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