2026届广州市九年级数学中考冲刺模拟试卷（含答案详解与评分标准）

2026届广州市九年级数学中考冲刺模拟试卷

（含答案详解与评分标准）学校：____________班级：____________姓名：____________考号：____________考试时间：120分钟满分：120分考试范围：九年级数学综合用途：中考冲刺检测注意事项：1．本试卷共三大题，22小题，满分120分，考试时间120分钟。请在规定位置填写学校、班级、姓名和考号。2．选择题每小题3分，共30分；填空题每小题3分，共18分；解答题共6题，共72分。3．作答时请写清必要的计算过程、推理依据和结论；书写不规范、跳步严重或结论与过程不一致的，按评分标准扣分。4．本卷用于考前冲刺综合检测，重点考查基础运算、函数图象、统计概率、几何推理和实际应用能力。选择题、填空题答题栏题号12345678910111213141516答案题号171819202122得分一、选择题（本大题共10小题，每小题3分，共30分。每小题只有一个正确选项）1．计算式的值是（）A．1B．3C．5D．72．2026年广州中考冲刺复习中，某校统计一次线上答题的平均用时为0.000035小时。用科学记数法表示0.000035，正确的是（）A．3.5×10⁻⁴B．3.5×10⁻⁵C．0.35×10⁻⁴D．35×10⁻⁶3．若a＜b，则下列结论一定成立的是（）A．a＋2＞b＋2B．－2a＜－2bC．a－3＜b－3D．a²＜b²4．不等式3x－4≤2(x＋1)的解集是（）A．x≤－6B．x≥－6C．x≤6D．x≥65．一次函数y＝－2x＋4与x轴的交点坐标是（）A．(0，4)B．(2，0)C．(－2，0)D．(0，－2)6．某冲刺小组6名同学一次限时训练成绩为7，8，8，10，12，15。该组成绩的中位数是（）A．8B．8.5C．9D．107．一个正方形的边长为6cm，则它的对角线长为（）A．6cmB．12cmC．6√2cmD．3√2cm8．在⊙O中，∠AOB＝100°，点C在优弧AB上，则∠ACB的度数为（）A．40°B．50°C．80°D．100°9．袋中有3个红球和2个白球，除颜色外完全相同。从中不放回地随机摸出2个球，恰好摸到1个红球和1个白球的概率是（）A．2/5B．3/5C．3/10D．4/510．二次函数y＝x²－4x＋1的最小值是（）A．－5B．－4C．－1D．－3二、填空题（本大题共6小题，每小题3分，共18分）11．分解并化简：(x＋3)²－(x－1)²＝__________。12．方程2x－5＝3的解为x＝__________。13．反比例函数y＝k/x的图象经过点(－3，2)，则k＝__________。14．在Rt△ABC中，∠C＝90°，AB＝10，sinA＝3/5，则BC＝__________。15．一个正多边形的每个外角都是40°，则这个正多边形的边数是__________。16．二次函数的图象开口向上，顶点为(1，－4)，且与y轴交于(0，－3)，则它的函数表达式为__________。三、解答题（本大题共6小题，共72分。解答应写出文字说明、证明过程或演算步骤）17．（12分）本题考查分式化简、代入求值和一元一次不等式组，属于考前运算稳定性检测。（1）先化简，再求值：（2）解不等式组，并写出它的所有整数解：作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（12分）某校九年级在中考冲刺阶段组织一次数学限时检测，随机抽取25名学生的成绩（满分100分）进行统计，整理如下表。成绩x（分）60≤x＜7070≤x＜8080≤x＜9090≤x≤100人数36124（1）求这25名学生中成绩不低于80分的人数所占百分比；（2）若全年级共有480名学生，估计本次检测成绩不低于80分的学生人数；（3）在成绩不低于90分的4名学生中，有1名男生、3名女生。若从这4人中随机抽取2人交流冲刺经验，求抽到的2人都是女生的概率。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（12分）如图，在△ABC中，D、E分别是AB、AC的中点，连接DE。已知BC＝8，△ABC中BC边上的高为6。（1）证明：DE∥BC；（2）求DE的长；（3）求四边形DBCE的面积。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（12分）已知抛物线y＝x²＋bx＋c经过点A(－1，0)、B(3，0)。直线l：y＝－x＋3与该抛物线交于点B和另一点C。（1）求抛物线的函数表达式和顶点坐标；（2）求点C的坐标；（3）求△ABC的面积。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（12分）某文具店销售“中考冲刺套装”，每套成本为30元。根据近两周销售数据，若每套售价为x元，日销售量y套满足y＝140－2x，且40≤x≤60。（1）用含x的代数式表示每日利润W，并化为一般式；（2）当售价为多少元时，每日利润最大？最大利润是多少元？（3）若要求每日利润不低于750元，且日销售量不少于40套，求售价x的取值范围。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22．（12分）如图，矩形ABCD中，AB＝12，AD＝9，连接BD。点P在AB上，设AP＝x（0＜x＜12），过点P作PE∥AD，交BD于点E。（1）用含x的代数式表示PE的长；（2）求△BPE的面积S关于x的函数表达式；（3）若△BPE与四边形APED的面积比为1∶3，求x的值。作答区：______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与解析评分标准说明：选择题每题3分，选对得3分，错选、多选或不选不得分；填空题每题3分，答案等价即可得分；解答题按过程给分，关键步骤正确但计算有轻微失误时酌情给分，结论与过程矛盾时以过程为准。答案汇总与评分要点速查题号12345678答案ABCCBCCB题号910111213141516答案BD8x＋84－669y＝(x－1)²－4题号主要考点核心给分点满分17分式化简、不等式组限制条件；通分约分；代入求值；解集与整数解1218统计估计、简单概率百分比；样本估计总体；列举等可能结果；概率1219三角形中位线、面积证明平行；求中位线；相似面积比或割补面积1220二次函数与一次函数求解析式；求交点；利用底和高求面积1221二次函数实际应用利润函数；最大利润；利润与销量条件联立1222矩形、相似、面积函数相似比例；面积表达式；面积比方程求解12题号小问分值得分细则17（1）6限制条件1分；通分2分；约分2分；代入1分17（2）6两个不等式各2分；求交集1分；整数解1分18（1）4不低于80分人数2分；百分比2分18（2）3列估计式2分；结果1分18（3）5列举或计数3分；有利结果1分；概率1分19（1）4中点关系2分；平行结论2分19（2）3中位线或相似比2分；长度1分19（3）5大三角形面积2分；面积比2分；结论1分20（1）5解析式3分；顶点坐标2分20（2）4联立方程2分；横坐标1分；坐标1分20（3）3底长2分；面积1分21（1）4利润关系式2分；一般式2分21（2）4对称轴2分；最大利润2分21（3）4利润不等式1分；范围2分；销量条件1分22（1）4相似关系2分；比例式1分；表达式1分22（2）4面积式2分；化简2分22（3）4面积关系1分；方程1分；求解1分；取值说明1分合计解答题72过程清楚、结论正确，按以上细则累计一、选择题答案与关键理由1．A。(－2)²＝4，√9＝3，所以原式＝4－3＝1。2．B。0.000035把小数点向右移动5位得到3.5，因此写成3.5×10⁻⁵。3．C。不等式两边同时减去同一个数，不等号方向不变，所以a－3＜b－3一定成立。4．C。3x－4≤2x＋2，移项得x≤6。5．B。与x轴交点的纵坐标为0，令－2x＋4＝0，得x＝2。6．C。数据已按从小到大排列，6个数的中位数是第3个和第4个数的平均数，即(8＋10)/2＝9。7．C。正方形对角线、两条边构成等腰直角三角形，对角线长为6√2cm。8．B。圆周角等于所对圆心角的一半，∠ACB＝100°/2＝50°。9．B。两球一红一白可按“先红后白”或“先白后红”计算，概率为3/5×2/4＋2/5×3/4＝3/5。10．D。y＝x²－4x＋1＝(x－2)²－3，平方项最小为0，所以最小值为－3。二、填空题答案与解析11．答案：8x＋8。解析：利用平方差：(x＋3)²－(x－1)²＝[(x＋3)－(x－1)][(x＋3)＋(x－1)]＝4(2x＋2)＝8x＋8。12．答案：4。解析：2x－5＝3，移项得2x＝8，所以x＝4。13．答案：－6。解析：反比例函数中k＝xy，将点(－3，2)代入得k＝－3×2＝－6。14．答案：6。解析：在Rt△ABC中，sinA＝BC/AB＝3/5，AB＝10，所以BC＝10×3/5