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2026年广东省初中学业水平模拟考试（二）·
数学
本试卷共4页，23小题，满分120分.考试用时120分钟.
注意事项：1.答题前，考生务必用黑色字迹的签字笔或钢笔将自己的学校、姓名和准考号填写在答题卡上
将条形码粘贴在答题卡“条形码粘贴处”，
2.作答选择题时，选出每小题答案后，用2B铅笔将答题卡上对应题目选项的答案信息点涂黑；
如需改动，用橡皮擦干净后，再选涂其他答案，答案不能答在试卷上：
3.非选择题必须用黑色字迹的签字笔或钢笔作答，答案必须写在答题卡各题目指定区域内的相应
位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液.不按
以上要求作答的答案无效.
4.考生必须保持答题卡的整洁，考试结束后，将试卷和答题卡一并交回
一、选择题：本大题共10小题，每小题3分，共30分.在每小题给出的四个选项中，只有一项
是符合题目要求的.
1.8的立方根是()
A.8
B.±2
C.2
D.-2
2.下列运算正确的是（)
A.2a·3a=6a
B.a5÷a2=a3
C.(a2)3=a5
D.（-ab)2=a2b2
3.如题3图，在Rt△ABC中，∠C=90°，BC=4,AC=3,则sinB=(
)
A
5
B.4
C.3
3
4
p.
-D
:E
A
O
D
B
A-------B
B
FQ P
C
题3图
题4图
题7图
题8图
4.如题4图是某个几何体的展开图，该几何体是()
A.三棱锥
B.三棱柱
C.四棱锥
D.
四棱柱
5.随着新一轮科技革命和产业变革逐步走向纵深，我国新能源汽车产业实现了快速发展，新能
源汽车已经成为我们日常出行的重要交通工具.据统计，截至2025年底，我国新能源汽车保
有量达4397万辆，其中“4397万”用科学记数法表示为()
A.0.4397×108
B.4.397×108
C.4.397×107
D.4.397×106
6.每周五下午的社团课是学校的特色课程，同学们可以选择自己喜欢的课程.小明和小丽从“篮
球课”“思辨课”“机器人课”三种课程中随机选择一种参加，则两人恰好选择同一种课程的
概率是()
A月
B.
c
D.
9
7.化学老师布置了“测量锥形瓶内部底面的内径”的任务.如题7图，小明将两根小棒AD,
BC的中点O固定，测得C,D之间的距离即内径AB的长度.此方案依据的数学定理是()
A.边角边
B.角边角
C.边边边
D.角角边
8.如题8图，一束光线PO从空气中斜射入长方体玻璃砖发生折射，已知AD/1BC,延长PO交
BC于点P',若∠POA=50°，∠P'OQ=25°，则∠OQB的度数为()
A.45°
B.55°
C.65
D.75°
数学第1页（共4页)2026年广东省初中学业水平模拟考试（二）
数学参考答案及评分标准
一、选择题：本大题共 10小题，每小题 3分，共 30分．在每小题给出的四个选项中，只有一项
是符合题目要求的．
题号 1 2 3 4 5 6 7 8 9 10
答案 C D D B C A A D A D
二、填空题：本大题共 5小题，每小题 3分，共 15分．
3
11 3．－ 5 12．－6 13． 14．＞ 15．32
三、解答题（一）（共 3小题，每小题 7分，共 21分）
16．解：（1）根据计算步骤可知，第②步是分式的通分，通分的依据是分式的基本性质，
故答案为：②，分式的基本性质；······························································2分
（2）不正确，第③步出现错误，正确的解题步骤如下：
16 2
原式＝ － ····································································
(m－4)(m＋4 m 4 3分) －
16 2(m＋4)
＝ － ···························································· 4分
(m－4)(m＋4) (m－4)(m＋4)
16－2m－8
＝ ··················································································· 5分
(m－4)(m＋4)
－2(m－4)
＝ ··················································································· 6分
(m＋4)(m－4)
2
＝－ ．···························································································m 4 7分＋
17．（1）8.5，8；································································································2分
（2）八；····································································································· 4分
（3）我认为九年级的学生体育锻炼情况的总体水平较好，因为八、九年级的平均数相等，九
年级每周锻炼时间的方差小于八年级每周锻炼时间的方差，所以九年级的学生体育锻炼
情况的总体水平较好．·············································································7分
18．解：（1）w关于 t的函数图象如图所示：
······························································· 2分
数学参考答案及评分标准 第 1页 （共 7页）
根据图象发现容器内盛水量 w（mL）与滴水时间 t（min）符合学习过的一次函数，
故答案为：一次；···················································································3分
（2）设一次函数解析式为 w＝kt＋b，将点（0，5），（5，20）代入得：
b＝5
，························································································· 4分
5k＋b＝20
k＝3
解得 ，·························································································· 6分
b＝5
∴一次函数解析式为 w＝3t＋5；································································ 7分
四、解答题（二）（共 3小题，每小题 9分，共 27分）
19．解：选①②为条件，③为结论组成一个命题，命题是真命题，理由如下：················ 3分
连接 BN，
∵四边形 ABCD是正方形，
∴∠C＝90°，∠CDM＝45°，···········································································4分
∵MN⊥BD，
∴∠NMB＝∠NMD＝90°，
∴△DMN是等腰直角三角形，········································································ 5分
∴DM＝NM，
在 Rt△BMN和 Rt△BCN中，
，
∴Rt△BMN≌Rt△BCN（HL），······································································· 7分
∴NM＝NC，
∴DM＝CN．······························································································· 9分
20．解：（1）设每台 A型机器人平均每小时清扫 x平方米，则每台 B型机器人平均每小时清扫
（x＋3）平方米．由题意，得 ，·················································· 1分
解得 x＝30，··························································································2分
经检验 x＝30是原方程的解，且符合题意，
∴x＋3＝30＋3＝33，·············································································· 3分
答：每台 A型机器人平均每小时清扫 30 平方米，每台 B型机器人平均每小时清扫 33
平方米．······························································································· 4分
（2）若物业公司共购进 20台机器人，A型机器人 2000元/台，B型机器人 3000元/台．公
司要求这批机器人每小时至少清扫 630平方米楼道，则：
设购进 n台 A型机，则购进（20－n）台 B型机．
由题意，得 30n＋33(20－n)≥630，··························································· 5分
解得 n≤10，··························································································6分
数学参考答案及评分标准 第 2页 （共 7页）
设总成本为 w元，则 w＝2000n＋3000(20－n)＝－1000n＋60000，···················7分
∵－1000＜0，n≤10，
∴当 n＝10时，总成本 w最低，································································ 8分
最低成本为：－1000×10＋60000＝50000，此时 20－n＝20－10＝10，
答：购买 10台 A型机，10台 B型机，能使总成本最低，总成本最低为 50000元．·9
分
21．解：（1）任务一：
如图 2，取圆的两弦 AB、CD，再分别作 AB和 CD的垂直平分线，两垂直平分线相交于点 O，
点 O为所作；
········································································· 3分
（2）任务二：
过 O点作 OH⊥CD于 H点，连结 OC、OD，如图，
∵正五边形 ABCDE内接于⊙O，
∴OC＝OD＝10cm，∠COD＝ ＝72°，······················································ 4分
∵OH⊥CD，
∴∠COH＝ ∠COD＝36°，CH＝DH，·····························································6分
在 Rt△OCH中，∵sin∠COH＝ ，
∴CH＝10×sin36°＝10×0.59≈5.9（cm），························································ 7分
∴CD＝2CH＝11.8cm．·················································································· 8分
答：内接正五边形 ABCDE的边长为 11.8cm．·····················································9分
数学参考答案及评分标准 第 3页 （共 7页）
五、解答题（三）（共 2小题，第 22题 13分，第 23题 14分，共 27分）
22．解：（1）∵y1＝ x，y2＝x＋y1，
∴y2＝x＋ x＝ x．················································································ 1分
∴y2的表达式为 y2＝ x；·········································································2分
（2）∵点 P（m，n）在函数 y1＝－3x＋2的图象上，
∴点 P的坐标为（m，－3m＋2）．
∵点 P关于函数 y1的“加和点”为点 Q，
∴点 Q的坐标为（m，－2m＋2）．····························································· 3分
∵Q与点 P的纵坐标互为相反数，
∴－3m＋2＋（－2m＋2）＝0．
解得：m＝ ．······················································································· 4分
∴－3m＋2＝－ ．
∴点 P的坐标为（ ，－ ）；·································································· 5分
（3）由（2）可得：点 Q的坐标为（ ， ），直线 QT的解析式为：y2＝－2x＋2．
当 x＝0时，y＝2，
∴点 T的坐标为：（0，2）．······································································ 6分
设直线 TP的关系式为 y＝kx＋2，
∴－ ＝ k＋2
解得：k＝－3．
∴直线 TP的解析式为：y＝－3x＋2．·························································7分
Ⅰ、①当正方形的顶点 A在直线 TP上时，四边形 ABCD的边与图形“M”有一个交
点．
∴－3t＋2＝t．
解得：t＝ ．························································································ 8分
②当正方形的顶点 A在直线 TQ上时，四边形 ABCD的边与图形“M”有 3个交点．
数学参考答案及评分标准 第 4页 （共 7页）
∴－2t＋2＝t．
解得：t＝ ．························································································ 9分
∴ ＜t＜ 时，四边形 ABCD的边与图形“M”有且只有 2个交点；·············· 10分
Ⅱ、①当正方形 ABCD的边 AD与 PQ重合时，四边形 ABCD的边与图形“M”有无数
个交点．
∴t＝ ．····························································································· 11分
②当正方形 ABCD的边 AB过点 T时，四边形 ABCD的边与图形“M”有 1个交点．
∴t＝2，
∴当 ＜t＜2．时，四边形 ABCD的边与图形“M有且只有 2个交点．············12分
数学参考答案及评分标准 第 5页 （共 7页）
故 t的取值范围是： ＜t＜ 或 ＜t＜2．················································· 13分
23．（1）解：∵长方形纸片 ABCD和 AQGF是两个完全相同的长方形，
∴∠BAC＝∠GAF，AC＝AG，
∴∠GAF＋∠CAD＝∠BAC＋∠CAD，
∴∠GAC＝∠BAD＝90°，·········································································1分
∴△ACG是等腰直角三角形，···································································2分
∴∠ACG＝45°，·····················································································3分
（2）①证明：∵BM＝AB，
∴∠BMA＝∠BAM，
∵AB//DC，
∴∠DMA＝∠BAM，··············································································· 4分
∴∠BMA＝∠DMA，
∴AM平分∠DMB．················································································ 5分
②过点 B作 BE⊥AF于点 E，
∵AD＝6，AF＝AB＝10，
∴DF＝ AF2－AD2＝8，
∴CF＝DC－DF＝2，·············································································· 6分
∵AF＝AB，
∴∠AFB＝∠ABF，·················································································7分
∵AB//DC，
∴∠CFB＝∠ABF，
∴∠CFB＝∠EFB，
又∵∠C＝∠BEF＝90°，BF＝BF，
∴△BCF≌△BEF（AAS），
∴CF＝EF＝2，BC＝BE，
∴AE＝AF－EF＝10－2＝8，·····································································8分
∵AD＝AQ＝BC，
∴AQ＝BE，
数学参考答案及评分标准 第 6页 （共 7页）
又∵∠AOQ＝∠EOB，∠QAO＝∠BEO＝90°，
∴△AOQ≌△EOB（AAS），
1
∴AO＝OE＝ AE＝4，············································································2 9分
（3）解：过点 F作 FH⊥BC交 BC延长线于点 H，
∵四边形 ABCD是正方形，
∴CB＝AB，∠ABC＝90°，
∴∠H＝∠ABC＝90°，
由旋转得 EF＝AE，∠AEF＝90°，
∵∠BEA＋∠BAE＝∠BEA＋∠FEH＝90°，
∴∠BAE＝∠FEH，
在△AEB和△EFH中，
∠BAE＝∠HEF
∠ABE＝∠H ，
AE＝EF
∴△AEB≌△EFH（AAS），
∴FH＝BE，AB＝EH，···········································································10分
∴EH＝CB，
∴CH＋CE＝CE＋BE，···········································································11分
∴CH＝BE，
∴FH＝CH，························································································ 12分
∴∠FCH＝45°，
∴∠BCG＝45°，··················································································· 13分
∵∠CBG＝90°，
∴△CBG是等腰直角三角形，
∴BG＝BC＝5 2．················································································ 14分
数学参考答案及评分标准 第 7页 （共 7页）

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