2023年中考押题预测卷01（苏州卷）-数学（参考答案）

2023年中考押题预测卷01【苏州卷】

数学·参考答案

9．

10．

11．2

12．或

13．/49度

14．

15．20/二十

16．/

17．【答案】0

【分析】根据二次根式的混合计算法则，零指数幂和负整数指数幂以及有理数的乘方等计算法则求解即可．

【详解】解：原式 ··············································1分

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

．···············································································5分

【点睛】本题主要考查了二次根式的混合计算，零指数幂，负整数指数幂，有理数的乘方，正确计算是解题的关键．

18．【答案】无解

【分析】分式方程去分母转化为整式方程，求出整式方程的解得到的值，经检验即可得到分式方程的解．

【详解】解：去分母得：，············································ 2分

解得：， ···································································· 3分

检验：当时，， ······················································· 4分

是增根，分式方程无解．·····················································5分

【点睛】此题考查了解分式方程，利用了转化的思想，把方程的解代入原方程进行检验是解题的关键．

19．【答案】化简：，值：

【分析】先利用完全平方公式与平方差公式进行乘法运算，再合并得到化简的结果，再把代入化简后的代数式进行计算即可．

【详解】解：

·······························································1分

；·············································································3分

当时，

原式．····························································6分

【点睛】本题考查的是整式的乘法运算，化简求值，熟练的利用平方差公式与完全平方公式进行简便运算是解本题的关键．

20．【答案】(1)

(2)

【分析】（1）先求出这组数中质数的个数，再利用概率公式解答即可；

（2）首先根据题意可直接列出所有可能出现的结果，再算出这个两位数能被3整除的概率.

【详解】（1）解：数字1，2，4，5中，2，5是质数，

则随机抽取1张，抽到卡片数字是奇数的概率为；

故答案为：；········································································2分

（2）解：列表如下：

出现的等可能性结果有12种，两位数能被3整除的有，

12，15，21，24，42，45，51，54共有8种，················································4分

∴P（两位数能被3整除）．························································6分

【点睛】本题考查的是用列表法或画树状图法求概率，列表法或画树状图法可以不重复不遗漏的列出所有可能的结果，适合于两步完成的事件，熟记概率公式是解决本题的关键．

21．【答案】(1)见解析

(2)

【分析】（1）求出，根据全等三角形的判定定理得出，推出，根据角平分线性质得出即可；

（2）证明，得到，利用得到，再利用线段的和差计算即可．

【详解】（1）

解：证明：，，

，

在和中，

，

，·························································2分

，

，，

平分；·····································································3分

（2）

在和中，

，

，······························································4分

∵，

∴，·········································································5分

，，

，

．··························································6分

【点睛】本题考查了角平分线的判定，全等三角形的性质和判定的应用，注意：全等三角形的判定定理有，，，，，全等三角形的对应边相等，对应角相等．

22．【答案】(1)，

(2)，

(3)人

【分析】（1）根据数据统计的方法以及各组数据之和等于样本容量可得答案；

（2）根据中位数、众数的定义可求出、的值；

（3）求出样本中甲乙两个班“优秀”所占的百分比，进而估计总体中“优秀”所占的百分比，再乘总人数即可．

【详解】（1）解：由题意可知，乙班在的数据有个，在的有，个，

故答案为：，；·······································································2分

（2）甲班人中得分出现次数最多的是分，共出现次，因此甲班学生成绩的众数，

将乙班名学生的成绩从小到大排列，处在中间位置的两个数的平均数为，因此中位数，

故答案为：，；·····································································4分

（3）（人），

答：甲班、乙班共人为样本估计全年级人中优秀人数约为人．·························8分

【点睛】本题考查中位数、众数，频数分布表，掌握中位数、众数以及“频率”是正确解答的前提．

23．【答案】(1)，

(2)或

【分析】（1）将、两点代入反比例函数解析式，，，可得，解得的值，即可求出、两点的坐标，用待定系数法即可求出一次函数和反比例函数的表达式；

（2）令，求出点坐标，根据、、三点坐标求出的面积，再得到的面积，设，利用三角形面积求出的值即可．

【详解】（1）由题意，得，解得，

，，

把代入，得，

反比例函数表达式为，·····························································2分

把，代入，得，

，

一次函数表达式为；···························································4分

（2）令，则得，，

点的坐标为，

，

，······································································5分

设，则，得，

，

解得：或，··································································7分

故或．································································8分

【点睛】本题考查了一次函数和反比例函数的综合应用，反比例函数与几何综合，待定系数法求函数解析式，三角形的面积的计算，正确求出一次函数和反比例函数解析式是解题的关键．

24．【答案】(1)见解析

(2)

【分析】（1）先连接交于，由切线的性质，平行线的性质，等腰三角形的性质推出，即可证明平分；

（2）平行线等分线段定理得到，推出是的中位线，求出，得到的长，由相似三角形的性质即可解决问题．

【详解】（1）

证明：连接交于，

∵切于，

∴，

∵，

∵，

∴，·······································································1分

∵，

∴，

∴，

∴平分；·····································································3分

（2）解：∵，

∴，

∵，

∴，

∴是的中位线，

∴，······································································5分

∵，

∴，

∴，····························································7分

∵，

∴．

∴的值是．·······································································8分

【点睛】本题考查切线性质，相似三角形的判定和性质，平行线的性质，角平分线的定义，三角形中位线定理，关键是连接求出的长．

25．【答案】(1)甲生产队每天需收割费元、乙生产队每天需收割费元．

(2)①甲乙两生产队分别工作的天数共有5种可能．②费用最少的方案是甲生产队工作5天，乙生产队工作34天，最低费用为元．

【分析】（1）设甲生产队每天需收割费x元、乙生产队每天需收割费元，根据当甲生产队所需收割费为5000元，乙生产队所需收割费为4000元时，两生产队工作天数刚好相同列出方程，解方程并检验即可；

（2）①设甲生产队工作m天，则乙生产队工作n天．由题意得到 ①，且 ②，由①得到③，整理得到关于m的一元一次不等式组，解得，由是正整数得到甲乙两生产队分别工作的天数共有5种可能．②得到总费用，根据一次函数的性质得到费用最少的方案和最少费用即可．

【详解】（1）解；设甲生产队每天需收割费x元、乙生产队每天需收割费元，

由题意，，··································································1分

解得，··········································································2分

经检验，是分式方程的解且符合题意．

则，

答：甲生产队每天需收割费元、乙生产队每天需收割费元．···························3分

（2）①设甲生产队工作m天，则乙生产队工作n天．

由题意， ①，且 ②，

由①得到③，

把③代入②得到，，即，·····················4分

∵，

∴，

∴，································································5分

解得，

∵是正整数，

∴或或或或．·········································7分

∴甲乙两生产队分别工作的天数共有5种可能．

②总费用，···········································8分

∵，

∴w随m的增大而增大，··································································9分

∴时，w有最小值，此时．

即费用最少的方案是甲生产队工作5天，乙生产队工作34天，最低费用为元．··············10分

【点睛】本题考查了分式方程的实际应用，一元一次不等式组的应用，一次函数的实际应用，考查较为全面，对于一次函数而言，当时，y随x的增大而增大，当时，y随x的增大而减小．

26．【答案】(1)，

(2)①；②

(3)存在，

【分析】（1）令，则，已知点的坐标，即可求得点的坐标，令，即可求得点的坐标；

（2）①由求得，根据平行线分线段成比例求解即可；②过点作轴交于点，求得直线的解析式，用含的式子表示，根据二次函数的性质即可得出结论；

（3）假设存在点使得，即，过点作轴交抛物线于点，由，可知平分，延长交轴于点，易证为等腰三角形，求得，则直线的解析式为，令，可得结论．

【详解】（1）解：令，则，

解得：或，

，

，

令，则，

，

故答案为：，；·······························································2分

（2）解：①轴，，，，

，

令，则，

解得：，或，

，·············································································3分

，

；·······································································4分

②如图，过点作轴交于点，

设直线的解析式为，

，，

，

解得，

直线的解析式为，·······················································5分

设点的横坐标为，则，则，

，

，

，

当时，的最大值为；·······················································6分

（3）解：假设存在点使得，即，过点作轴交抛物线于点，延长交x轴于点M，如图，

，，

，

轴，

，

，

为等腰三角形，·································································7分

，

，，

，············································································8分

设直线的解析式为，

，

，

解得，

直线的解析式为，······················································9分

令，

解得：或（与点重合，舍去），

存在点满足题意，此时．·······················································10分

【点睛】本题是二次函数综合题，考查了待定系数法求函数解析式，平行线分线段成比例，二次函数的性质，角度的存在性相关内容，熟练掌握知识点并灵活运用是解题的关键．

27．【答案】(1)见解析

(2)①3；②或

【分析】（1）利用证明，再由等量代换证明；

（2）①在上截取，连接交于点，过点作交于点，由（1）可知，则，再由平行线的性质可得，即设，，则，由，可得，，从而得到等式，求出，即可求；②延长至，使，连接交于点，过点作交于点，由（1）可知，则，可得，设，，则，可知，再由，分别得到，，从而得到方程，求出或，即可求或．

【详解】（1）解：证明：是等边三角形，

，，································································1分

，

，···································································2分

，

；············································3分

（2）①在上截取，连接交于点，过点作交于点，

由（1）可知，

，

，

，，

，·········································································4分

，

，

设，，则，

，

，即

，

，

，

，·········································································5分

解得或（舍），

；·········································································6分

②延长至，使，连接交于点，过点作交于点，

由（1）可知，

，

，·········································································7分

设，，则，

，

，

，

，，

，，

····································································8分

解得或，

或．·······································································10分

【点睛】本题考查了等边三角形的性质，三角形全等的判定及性质，平行分线段，熟练掌握性质定理是解题的关键．