2026届广州市九年级数学中考二模模拟试卷（含答案详解与评分标准）

2026届广州市九年级数学中考二模模拟试卷（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分考试节点：中考二模适用对象：九年级注意事项：1．本卷用于2026届广州市九年级数学中考二模考前综合检测，试题覆盖数与式、方程与不等式、函数、统计与概率、图形与几何、综合应用等核心内容。2．全卷共22题，其中选择题10题，每题3分，共30分；填空题6题，每题3分，共18分；解答题6题，共72分。满分120分，考试时间120分钟。3．答题前，请将学校、班级、姓名、考号填写清楚。选择题答案写在指定位置，填空题只写最终结果，解答题必须写出必要的文字说明、计算过程或证明过程。4．作图题可先用铅笔作图，再用黑色签字笔描清；答案中的根式、分式、π等保留准确值。不得使用计算器。5．交卷前请检查题号、答案和单位是否对应，避免因书写不清、步骤缺失或结果化简不规范造成失分。题型题量分值说明选择题、填空题、解答题10题、6题、6题30分、18分、72分合计120分一、选择题（本大题共10小题，每小题3分，共30分）每小题给出的四个选项中，只有一项是符合题目要求的。请把正确选项填写在题后括号内。1．在数轴上，数－2026的相反数是（）A．－2026B．2026C．1/2026D．－1/20262．下列运算正确的是（）A．a³＋a²＝a⁵B．(a²)³＝a⁶C．a⁶÷a³＝a²D．(ab)²＝a²＋b²3．某校九年级数学二模限时训练后，抽取8名学生的成绩为45，48，48，49，50，50，50，51，则这组数据的中位数是（）A．48B．49C．49.5D．504．如图意：两条直线a，b被同一条截线所截，若a∥b，且其中一个内错角为68°，则与它相邻的同旁内角的度数是（）A．68°B．78°C．102°D．112°5．一元二次方程x²－5x＋6＝0的两个根分别是（）A．－2，－3B．1，6C．2，3D．－1，－66．二次函数y＝(x－2)²＋1的最小值是（）A．－2B．1C．2D．不存在7．一个不透明袋中有3个红球和2个白球，这些球除颜色外完全相同。随机摸出一个球，放回摇匀后再摸出一个球，两次摸出的球颜色相同的概率是（）A．6/25B．12/25C．13/25D．3/58．在Rt△ABC中，∠C＝90°，AB＝10，sinA＝3/5，则AC的长为（）A．8B．6C．5D．49．点P在反比例函数y＝k/x（x＞0）的图象上，过点P分别作x轴、y轴的垂线，垂足为A，B。若矩形OAPB的面积为12，则k的值为（）A．－12B．6C．12D．2410．已知二次函数y＝x²－2mx＋m²－1的图象与x轴的两个交点都在0到4之间（含端点），则m的取值范围是（）A．0≤m≤4B．1≤m≤3C．－1≤m≤5D．2≤m≤4二、填空题（本大题共6小题，每小题3分，共18分）请把答案填写在题中横线上。答案须化为最简形式，带单位的题目应写明单位。11．用科学记数法表示0.0002026，结果为________________。12．分解因式：3a²－12＝________________。13．若一组数据2，5，6，x，9的平均数为6，则x＝________________。14．半径为3、圆心角为120°的扇形面积为________________。15．在平面直角坐标系中，点A(1，2)绕原点逆时针旋转90°后得到点A′，则A′的坐标是________________。16．如图意：正方形ABCD的边长为6，点E是CD的中点，连接AE，则点B到直线AE的距离为________________。三、解答题（本大题共6小题，共72分）解答应写出必要的文字说明、证明过程或演算步骤。各题保留的作答区用于正式考试书写，评分时按步骤给分。17．（本题10分）完成下列计算与化简。（1）计算：（2）先化简，再求值：作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（本题10分）为了解九年级学生完成一套数学二模限时训练所用时间t（单位：分钟）的情况，某区从参加训练的学生中随机抽取50名，整理得到如下分组统计表：时间t/分钟70≤t＜8080≤t＜9090≤t＜100100≤t＜110110≤t＜120人数41016128频数占比组中值758595105115（1）补全表中的“频数占比”（用百分数表示）；（2）若全区共有2800名九年级学生参加此类限时训练，请估计用时在90≤t＜110分钟的学生人数；（3）现从90≤t＜100分钟组中选出4名学生代表A，B，C，D，再随机抽取2名交流复习方法。请用列表或画树状图的方法，求“至少抽到学生A”的概率。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（本题12分）如图，在△ABC中，AB＝AC，D是BC的中点，E在AB上，F在AC上，且BE＝CF。（1）求证：△ADE≌△ADF，并说明ED＝FD；（2）若AB＝AC＝10，BC＝12，BE＝CF＝5，求△DEF的面积。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（本题12分）二模复习期间，广州某文创店销售“数学错题本”。每本进价为8元，若每本售价为x元（12≤x≤20），每周销量y（本）与售价x满足一次函数关系：y＝－20x＋500。设每周利润为w元。（1）求w关于x的函数表达式；（2）若售价可以按0.5元为单位调整，求每本售价为多少元时每周利润最大，并求最大利润；（3）若要求每周销量不少于160本，且每周利润不少于1400元，求售价x的取值范围。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（本题14分）如图，⊙O的直径AB＝10，点C在⊙O上，AC＝6，BC＝8。过点C作⊙O的切线，与直线AB交于点D（D在A的左侧）。（1）求证：∠ACD＝∠ABC；（2）求CD的长；（3）点E在射线DC上，且DE＝DB，连接BE，求△BCE的面积。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22．（本题14分）如图，抛物线y＝ax²＋bx＋c经过A(－1，0)、B(3，0)、C(0，3)三点，O为坐标原点。（1）求抛物线的函数表达式；（2）点P为抛物线在第一象限内的一动点，过P作PD⊥x轴，垂足为D，连接OP。设P(t，－t²＋2t＋3)，0＜t＜3，求△OBP面积的最大值及此时t的值；（3）直线l：y＝－2x＋m与抛物线交于M，N两点。若MN＝2√5，求m的值，并判断直线l是否经过△ABC的内部；若不经过，请说明它与△ABC的公共点情况。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与解析说明：本部分给出逐题答案、关键理由与评分标准。选择题、填空题按最终答案给分；解答题按步骤给分，学生使用等价方法、推理完整且结论正确的，可参照相同得分点给分。题号12345678910答案BBCDCBCACB分值3333333333选择题关键理由：1．相反数是只有符号相反、绝对值相等的两个数，因此－2026的相反数为2026。2．幂的乘方应把指数相乘，(a²)³＝a⁶；A项不是同类项合并，C项应为a³，D项应为a²b²。3．8个数据按从小到大排列后，第4个和第5个数分别为49、50，中位数为(49＋50)/2＝49.5。4．两直线平行时，内错角相等；同旁内角互补，因此相邻的同旁内角为180°－68°＝112°。5．x²－5x＋6＝(x－2)(x－3)，令每个因式为0，得两个根为2和3。6．y＝(x－2)²＋1中平方项的最小值为0，所以函数最小值为1。7．放回摸球两次相互独立，同色概率为(3/5)²＋(2/5)²＝9/25＋4/25＝13/25。8．sinA＝BC/AB＝3/5，AB＝10，得BC＝6，再由勾股定理得AC＝8。9．反比例函数第一象限图象上点P构成的矩形OAPB面积等于k，因此k＝12。10．该函数可写为y＝(x－m)²－1，两个零点为m－1和m＋1；两交点均在[0，4]内，得0≤m－1且m＋1≤4，所以1≤m≤3。题号111213141516答案2.026×10⁻⁴3(a－2)(a＋2)83π(－2，1)12√5/5分值333333填空题解析：11．把0.0002026的小数点向右移动4位得到2.026，因此0.0002026＝2.026×10⁻⁴。12．先提取公因式3，得3(a²－4)，再利用平方差公式得3(a－2)(a＋2)。13．由平均数定义可得(2＋5＋6＋x＋9)/5＝6，故22＋x＝30，x＝8。14．扇形面积为圆面积的120°/360°，所以S＝120/360×π×3²＝3π。15．点(x，y)绕原点逆时针旋转90°后为(－y，x)，故A(1，2)变为A′(－2，1)。16．设A(0，0)，B(6，0)，D(0，6)，C(6，6)，则E(3，6)，直线AE方程为y＝2x，即2x－y＝0。点B到该直线的距离为|12|/√5＝12√5/5。三、解答题参考答案与评分标准17．答案：（1）原式＝3√2－(√2－1)＋4－√2＝3√2－√2＋1＋4－√2＝5＋√2。解析要点：√18应先化为3√2；因为1－√2＜0，所以|1－√2|＝√2－1；(1/2)⁻²＝4，2cos45°＝√2。四项合并时要把常数项与根式项分别整理。（2）原式＝[(x－2)(x＋2)]/(x＋2)²÷[(x－2)/(x＋2)]。当x≠±2时，除以一个分式等于乘以它的倒数，所以原式＝[(x－2)(x＋2)]/(x＋2)²×[(x＋2)/(x－2)]＝1。把x＝2＋√3代入，满足x≠±2，故结果为1。得分点分值第（1）问正确化简√18、绝对值、负指数幂和三角函数值3分第（1）问合并同类根式并得出5＋√22分第（2）问正确分解因式并把除法转化为乘法3分第（2）问说明取值有效并得出结果12分18．答案：（1）频数占比依次为8%，20%，32%，24%，16%。计算方法为各组人数除以总人数50，例如90≤t＜100组的占比为16÷50＝32%。（2）用时在90≤t＜110分钟的抽样人数为16＋12＝28，占比为28/50＝56%。估计全区人数为2800×56%＝1568（人）。（3）从A，B，C，D中任取2名，共有AB，AC，AD，BC，BD，CD六种等可能结果，其中至少抽到A的结果有AB，AC，AD三种，故所求概率为3/6＝1/2。得分点分值补全5个频数占比，格式可为百分数或等值小数3分正确确定90≤t＜110分钟的频数和占比2分正确估计全区人数1568人2分列表或列举等可能结果完整，并得出概率1/23分19．答案：（1）因为AB＝AC，D是BC的中点，所以AD是底边BC上的中线，也是顶角平分线，即∠BAD＝∠DAF。又BE＝CF，且AB＝AC，所以AE＝AB－BE，AF＝AC－CF，从而AE＝AF。再加上AD＝AD，依据SAS可得△ADE≌△ADF。因此对应边ED＝FD。（2）由AB＝AC＝10，BC＝12，D为BC中点，得BD＝DC＝6。在Rt△ABD中，AD＝√(AB²－BD²)＝√(100－36)＝8。因为BE＝CF＝5且AB＝AC＝10，所以E，F分别是AB，AC的中点。由三角形中位线定理，EF∥BC且EF＝BC/2＝6。E，F到BC的距离等于A到BC距离的一半，即为4，所以D到EF的距离为4。故S△DEF＝1/2×EF×4＝12。得分点分值由等腰三角形性质得出∠BAD＝∠DAF或AD垂直平