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5.右图是一块积木及其主视图，则它的左视图是
正面
主视图
数
学
B
D
(第5题图)
6.将分式方程x-2.3
=1去分母后得到的整式方程为
注意事项：
2x-11-2x
1.本试卷分第I卷和第Ⅱ卷两部分.全卷共8页，满分120分，考试时间120分钟
A.x-2+3=2x-1
B.x-2+3=1
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
C.x-2-3=2x-1
D.x-2-3=1
3.答案全部在答题卡上完成，答在本试卷上无效，
7.如图，已知反比例函数y=6
(x>O)的图象经过点A,连
4.考试结束后，将本试卷和答题卡一并交回
接OA.将线段OA绕点A逆时针旋转，当点O的对应点O'
第I卷选择题（共30分）》
落在x轴上时，△OAO'的面积是
(第7题图)
A.3
B.6
C.9
D.12
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，
8.词元(Token)是大模型处理信息的最小信息单元，具有智能时代可计量、可定价、
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.某型号汽油每升的价格上涨0.8元记作“+0.8元”，则“-0.5元”表示这种汽油每
可交易的特征.2026年3月，中国日均词元调用量已突破1.4×104 Token.已知每消
升的价格
耗一度电大约可产出5.6×10 Token.由此估计，产出1.4×104 Token所消耗的电量
A.下降0.5元
B.上涨0.5元
用科学记数法表示为
C.下降0.3元
D.上涨0.3元
A.0.25×108度
B.2.5×10度
2.某校音乐爱好者成立了一支名为“火红”的乐队，并以乐队名首字母“”为元素
C.2.5×10度
D.25×10°度
设计了如下四种备选队徽图案，其中是中心对称图形的是
9.某国产机车工厂生产仿赛车与复古街车两种车型.已知生产1台仿
赛车比生产1台复古街车的成本高0.5万元，且生产5台仿赛车与
生产6台复古街车的成本相等.设生产1台仿赛车的成本为x万元，
生产1台复古街车的成本为y万元，则可列方程组为
3.计算（-x)2·x的结果是
x-y=0.5,
x-y=0.5,
y-x=0.5,
y-x=0.5,
A.ro
B.x8
C.-x5
D.-x8
A.
B.
C.
9
6x=5y
5x=6y
|5x=6y
6x=5y
4.2026年4月15日，山西五大文脉旅游线路发布，分别是：华夏之根、土木华章
10.如图，已知平面直角坐标系中，点A的坐标为(3,0)，点B的坐
晋魂春秋、雄关万里、表里山河.某自媒体创作者计划从这五条线路中随机选择
标为(m,n).若△AOB的面积为6，则下列说法一定正确的是
条进行实地探访，则他选中“华夏之根”线路的概率为
A.m=±2，n为任意实数
B.m=±4，n为任意实数
1
2
A.25
B.10
1
C.5
D.5
C.m为任意实数，n=±2
D.m为任意实数，n=±4
(第10题图)
数学第1页（共8页）
数学第2页（共8页）数 学
一、选择题（每小题 3分，共 30分）
题号 1 2 3 4 5 6 7 8 9 10
答案 A C A C D A B C B D
二、填空题（每小题 3分，共 15分）
11．2 12 12．70 13．14.5 14．2 15．
11
三、解答题（共 75分）
16.（每小题 5分，共 10分）
1
解：（1）原式 8 （ 9） 1 ········································································ 3分
3
8 3 1 ················································································ 4分
6 .·························································································5分
2 1 a 1 2（ ）原式 ·························································7分
a 2 (a 2)(a 2) a 1
1 (a 2)(a 2) 2 ···························································· 8分
a 2 a 1 a 1
a 2 2
·············································································9分
a 1 a 1
a
.··················································································· 10分
a 1
17．（本题 6分）
解：连接 OD.······························································································· 1分
∵ O与 BC相切于点 B，
∴ AB BC .
∴ ABC 90 .········································2分
∵ C 35 ，
∴ A 90 C 55 .·············································································3分
∵ BOD是 所对的圆心角， A是 所对的圆周角，
∴ BOD 2 A 110 .·············································································4分
∵AB=4，
∴OB=2.································································································ 5分
∴ 的长= 110 π 2 11 π .······································································ 6分
180 9
数学答案 第 1 页（共 5 页）
18．（本题 8分）
解：（1）165.5，164，3.25，A；······································································5分
【评分说明：方差 m的值 2 分，其余各空均为 1分】
（2）平均数不变；······················································································· 6分
方差发生变化，且方差变小.··································································· 8分
19．（本题 8分）
解：（1）设可购买 件立体拼图．·····································································1分
根据题意，得 88 + 38 80 6000. ············································· 4分
296解，得 .·············································································· 5分
5
因为 为整数，且 取最大值，所以 = 59.············································· 6分
答：最多可购买立体拼图 59件．·······························································7分
（2）13．··································································································· 8分
20.（本题 8分）
解：如图，延长 BM交 l1于点 P，延长 CN交 l1于点 Q.·········································1分
由题意得，四边形 BCQP是矩形.
∴PQ=BC=60米，BP=CQ.················· 2分
在 Rt△APB中，∠APB=90°，∠ABP=26.6°，
∴tan∠ = .

∴ PB PA PA PA 2PA. ···················································· 3分
tan∠ABP tan 26.6 0.5
∴CQ=2PA.
在 Rt△AQC中，∠AQC=90°，∠ACQ=58°，
∴tan∠ = .

∴QA QC tan ACQ 2PA tan58 3.2PA. ·················································· 4分
∵PA+QA=PQ，
∴PA+3.2PA=60.······················································································ 5分
解，得 PA≈14.3. ····················································································6分
∴PB=2PA= 2 14.3 28.6 29 米 . ··························································· 7分
答：这段河道的宽约为 29米.·········································································· 8分
数学答案 第 2 页（共 5 页）
21．（本题 9分）
解：（1）① 1；···························································································· 2分
②∵四边形 ABCD为菱形，
∴OA OC 1 AC，OB OD 1 BD， AC BD .············································ 3分
2 2
OA 3
∴ ，∠AOB=90°.··········································································· 4分
OB 3
∵在 Rt△AOB OA中， tan ABD ，
OB
∴ tan ABD 3 .·······················································································5分
3
∴ ABD 30 .··························································································· 6分
（2）答案不唯一，例如：
作法一： 作法二：
N M N M
作法三： 作法四：
················································································································· 8分
如图，菱形 ABMN即为所求.··········································································· 9分
数学答案 第 3 页（共 5 页）
22.（本题 13分）
解：（1）将 x=1，y=25和 x=2，y=40分别代入关系式 y ax2 v0x，得
25 a v0 ,
··········································································· 2分
40 4a 2v0.
解这个方程组，得
a 5,
··················································································· 4分
v0 30.
所以，y与 x的关系式为 y 5x2 30x .·························································5分
（2）①50 2.5x；······················································································· 6分
②由题意，得 5(x 3)2 30x 50 2.5x .······················································· 8分
化简，得 2x2 13x 20 0 .
5
解，得 x1 ， x2 4 .············································································· 10分2
5
答：小球甲与无人机在空中离地面的高度恰好相等时，x的值为 或 4.
2
3 13 19（ ） 或 .···························································································· 13分
4 4
【评分说明：只写出一个正确答案得 2 分，全部正确得 3 分】
23.（本题 13分）
解：（1） HG HE .························································································1分
理由：连接 CH. ·······················································································2分
∵四边形 ABCD是矩形，
∴ B 90 .
由折叠可知 FG BE， F B 90 ,CF CB .········3分
又∵CH=CH，
∴Rt△CFH≌Rt△CBH.
∴FH=BH.································································································4分
∴ FH FG BH BE，
即 HG HE .·····························································································5分
（2）①CG BE GM .··················································································6分
理由：由折叠可知 F B 90 ，CG CE， FG BE， FCG BCE .·········· 7分
数学答案 第 4 页（共 5 页）
∵四边形 ABCD是矩形，
∴ DCB 90 .
∴ BCE DCE 90 .
∴ FCG DCE 90 .
∴ FCE 90 .
∵ NEC 90 ，
∴四边形 FCEM是矩形.············································································· 8分
∴CE FM .·····························································································9分
∵ FM FG GM ,
∴CG BE GM .·····················································································10分
3 5 5（ ）G，N两点之间的距离为 5 或 .······················································ 13分
2
【评分说明：只写出一个正确答案得 2 分，全部正确得 3 分】
数学答案 第 5 页（共 5 页）

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