2027南京中考数学暑假衔接A+B双卷

—PAGE—原创可编辑交付版NANJINGMATHA+BPACK2027南京中考数学暑假衔接A+B双卷（基础诊断｜函数几何提升｜答案详解｜评分细则）A卷保留完整120分诊断与详细解析；B卷新增60分函数几何提升，用于暑假订正后的迁移再测。适合南京2027届初三开学分层、家庭自测和机构衔接班。01A卷120分完整诊断02B卷60分专题提升03评分·讲评·复盘2026暑假—2027南京中考适用可编辑·可打印·可复用

STARTHERE｜A卷定位，B卷验证先用A卷完成全科基础诊断；订正后，不要立刻再做同题，而应完成B卷函数几何提升，检验解析式、坐标、相似、圆、二次函数和最值能否迁移。A卷与B卷分值不同，请换算百分制比较。阶段时间产出A卷闭卷120分钟总分与模块得分逐题订正1—2天每题第一处错误+正确方法B卷再测60分钟函数几何迁移得分7天复测30分钟只重做两卷错题，检查稳定性!南京口径南京市2026年中考政策中数学满分为120分。本A卷采用120分制；2027年度正式安排如有变化，应以南京市教育部门最新发布为准。

2027南京中考数学暑假衔接A+B双卷（基础诊断｜函数几何提升｜答案详解｜评分细则）｜A卷：120分基础诊断2027届九年级数学一轮基础诊断A卷学校：________________班级：________姓名：________考号：________考试时间：120分钟满分：120分题型选择题填空题解答题总分分值301872120得分注意事项：1．本卷共三大题，22小题。请先检查试卷是否完整，再开始作答。2．选择题答案请填在答题位置；解答题须写出必要的文字说明、证明过程或演算步骤。3．本卷可直接打印训练，作答时保持书写清楚，合理安排时间。4．参考答案与解析另起新页，训练时建议先独立完成，再对照订正。一、选择题（本大题共10小题，每小题3分，共30分。每小题只有一个选项符合题意。）1．数−3/2的倒数是（）。A．3/2B．−3/2C．2/3D．−2/32．南京某场馆年度接待人数约为20260人，用科学记数法表示20260，正确的是（）。A．2.026×10³B．2.026×10⁴C．20.26×10³D．0.2026×10⁵3．若a≠b，则代数式(a²−b²)/(a−b)化简的结果是（）。A．a−bB．a+bC．a²+b²D．14．已知一个正方体的表面积为96cm²，则它的体积为（）。A．64cm³B．48cm³C．36cm³D．24cm³5．一元二次方程x²−4x+1=0的两个根分别为x₁，x₂，则x₁+x₂的值为（）。A．−4B．−1C．1D．46．一次函数y=kx+b的图像经过点(0，3)和(2，−1)，则它与x轴的交点横坐标是（）。A．1/2B．1C．3/2D．27．一个不透明袋中有2个红球、1个黄球、1个白球，除颜色外完全相同。从中不放回地随机摸出2个球，摸出的两个球恰好是1个红球和1个黄球的概率为（）。A．1/6B．1/4C．1/3D．1/28．点A、B、C在同一圆上，O为圆心。若∠AOB=112°，且点C在不含点A、B的较大圆弧上，则∠ACB的度数是（）。A．34°B．56°C．68°D．112°9．抛物线y=x²−2mx+m²−1与x轴交于两点，则这两点之间的距离为（）。A．1B．2C．2mD．与m的取值有关10．若整数x同时满足2x−3≤x+2<3x+5，则这样的整数x的个数为（）。A．5B．6C．7D．8二、填空题（本大题共6小题，每小题3分，共18分。请把答案直接写在横线上。）11．计算：√12−√3=________。12．分解因式：x²−6x+9=________。13．不等式组x−1≥0，2x−3<5的解集是________。14．点P(2，−3)关于x轴的对称点坐标为________。15．一个扇形的弧长为4π，圆心角为120°，则该扇形的半径为________。16．若直线y=2x+b与抛物线y=−x²+4x+5只有一个公共点，则b=________。三、解答题（本大题共6小题，共72分。请写出必要的文字说明、证明过程或演算步骤。）17．（10分）计算与解方程。（1）先化简，再求值：(a+2)²−a(a+4)，其中a=−3；（2）解分式方程：作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（10分）某校九年级开展“诊断考前限时训练”。数学组随机抽取40名学生一次训练成绩（满分100分），按分数段整理如下表：分数段60≤x<7070≤x<8080≤x<9090≤x≤100人数612166组中值65758595（1）估计这40名学生本次训练成绩的平均数；（2）若把80分及以上记为“达标”，估计该校九年级600名学生中达标人数；（3）从成绩低于70分的6名学生中随机选取2名参加一次面批，其中男生4名、女生2名，求选到的2名学生恰好1男1女的概率。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（12分）在矩形ABCD中，AB=4，BC=6，点E是BC的中点，连接AE。过点D作DF⊥AE，垂足为F。（1）求AE的长；（2）求点D到直线AE的距离DF；（3）若点P在AE上，且AP:PE=2:3，求△DPE的面积。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（12分）某校准备定制一批校园文化衫。市场调研发现：当每件售价为50元时，每周可售出600件；售价每提高1元，每周销量减少5件。已知每件成本为30元，设每件售价为x元（50≤x≤130），每周利润为W元。（1）用含x的式子表示每周销量，并写出W关于x的函数关系式；参考模型：（2）若每周利润不少于20000元，求售价x的取值范围；（3）学校希望在利润最大的同时尽量降低学生购买单价，应定价多少元？最大周利润是多少元？作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（14分）圆O的直径AB=8，点C在圆O上。过点C作圆O的切线，与AB的延长线交于点D，且∠D=30°（这里∠D指直线DC与直线DA所成的锐角）。（1）求证：△ACB为直角三角形；（2）求线段OD、CD的长；（3）求BC、AC的长，并判断△ABC的三个内角度数。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22．（14分）已知抛物线经过A(−1，0)、B(3，0)、C(0，3)三点。（1）求该抛物线的函数表达式和顶点坐标；目标形式：（2）过点C作直线y=kx+3，与抛物线的另一个交点为E。若点E在x轴上方，且△OBE的面积为3，求k的值；（3）点M在x轴上，点N在该抛物线上。若四边形ACNM为平行四边形，求点M、N的坐标。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析一、选择题答案与解析题号12345678910答案DBBADCCBBC1．D。倒数是乘积为1的两个数，−3/2的倒数为−2/3。2．B。20260=2.026×10⁴，科学记数法中a满足1≤a<10。3．B。a²−b²=(a−b)(a+b)，又a≠b，所以原式=a+b。4．A。设正方体棱长为s，则6s²=96，s²=16，s=4，体积s³=64cm³。5．D。由根与系数关系，x₁+x₂=−(−4)/1=4。6．C。由两点得k=(−1−3)/(2−0)=−2，故y=−2x+3；令y=0，得x=3/2。7．C。4个球中任取2个共有C₄²=6种；取1红1黄有2×1=2种，概率为2/6=1/3。8．B。同弧所对圆周角等于圆心角的一半。∠ACB所对的是弧AB，故∠ACB=112°÷2=56°。9．B。y=x²−2mx+m²−1=(x−m)²−1，与x轴交点横坐标为m−1和m+1，距离为2。10．C。由2x−3≤x+2得x≤5；由x+2<3x+5得x>−3/2。整数x为−1，0，1，2，3，4，5，共7个。1．易错点在于把“倒数”和“相反数”混淆。倒数要求分子分母互换并保留符号，负数的倒数仍为负数；相反数才是只改变符号。2．科学记数法的规范形式是a×10ⁿ，其中1≤a<10，指数n表示小数点移动的位数。20260从右向左移动4位得到2.026，因此指数为4。3．本题考查平方差公式与分式约分。约分前必须先说明a−b不为0，不能把a²−b²误看成(a−b)²，也不能在分母为0的情形下进行约分。4．正方体表面积是6个全等正方形面积之和。先由表面积求棱长，再求体积，步骤不能倒置；若直接把96开平方当作棱长，会忽略6个面的关系。5．根与系数关系适用于标准形式ax²+bx+c=0且a≠0。这里a=1，b=−4，两个根的和为−b/a，题目只问和，不需要求出两个根。6．两点确定一条直线。先求斜率，再利用点(0，3)确定截距。求与x轴交点时令y=0，而不是令x=0；x=0对应的是与y轴交点。7．不放回摸球适合用组合计数。先算总情况数，再算有利情况数。两个红球是不同的实物但颜色相同，计数时仍应按“从2个红球中选1个”处理。8．圆心角和圆周角的关系取决于它们是否对同一条弧。点C在较大圆弧上，∠ACB所对的是较小弧AB，仍等于对应圆心角的一半。9．把抛物线配方成(x−m)²−1后，零点来自(x−m)²=1。两个交点横坐标关于x=m对称，距离与m无关，这体现了二次函数平移不改变根间距的特征。10．连写不等式应拆成两个不等式分别求解，再取公共部分。最后要求整数个数，必须把端点是否能取纳入判断；本题左端是严格大于−3/2，右端是小于等于5。1．本题可在草稿纸上写出等式(−3/2)×□=1，再反推出空格中的数。若学生写成2/3，说明只做了分子分母互换，没有保留负号；若写成3/2，则把倒数错当成了绝对值。2．写科学记数法时，先确定有效数字部分，再确定10的幂。20260不是2.026×10³，因为2.026×1000=2026；也不写成20.26×10³，因为20.26不满足a的范围。3．分式化简的本质是因式约分。解题时先把分子因式分解，不要急于把a和b代入特殊值验证。若a=b，原式无意义，题干已给a≠b，正是为了保证约分合法。4．立体几何计算要分清“面积单位”和“体积单位”。表面积96cm²对应棱长4cm，体积才是64cm³。若选48或36，通常是把表面积平均分后没有继续求体积。5．根与系数关系在中考中常用于避免繁琐求根。此题判别式Δ=16−4=12>0，方程确有两个实根，但求和时并不需要写出含根号的根。6．一次函数图像经过(0，3)，说明截距b=3。再由另一个点求出k=−2，因此函数随x增大而减小。x轴交点是函数值为0的位置，横坐标为3/2。7．如果用树状图，也应体现第一次摸到球后第二次球的总数变为3。因为是不放回，不能把每次总数都当作4。组合方法更简洁，但前提是每种两球组合等可能。8．圆周角定理的关键是“同弧或等弧”。若把点C所在弧判断错，可能得到124°的一半或其他错误结果。题干明确C在不含A、B的较大圆弧上，使∠ACB对的是较小弧AB。9．参数m只控制抛物线左右平移，不改变开口大小和最低点到x轴的距离。零点关于对称轴x=m对称，两个零点的横坐标差为2，这是配方法的直接结论。10．本题是连写不等式的整数计数。公共解集为−3/2<x≤5，整数从−1开始到5结束。计数时可列举，也可用5−(−1)+1=7，注意不能把−2计入。二、填空题答案与解析题号111213141516答案√3(x−3)²1≤x<4(2，3)6611．√12−√3=2√3−√3=√3。12．x²−6x+9是完全平方公式，结果为(x−3)²。13．x−1≥0得x≥1；2x−3<5得x<4，故解集为1≤x<4。14．关于x轴对称，横坐标不变，纵坐标变为相反数，故为(2，3)。15．120°=2π/3，弧长l=rθ，所以4π=r·2π/3，r=6。16．联立得−x²+4x+5=2x+b，即x²−2x+(b−5)=0。只有一个公共点，则判别式Δ=4−4(b−5)=0，解得b=6。11．二次根式化简要先把被开方数分解成平方因数与剩余因数的乘积。√12=√(4×3)=2√3，之后再合并同类二次根式。12．完全平方三项式的结构是首项平方、末项平方、中间项为首末两项乘积的2倍。x²−6x+9中，9=3²，−6x=−2·3·x。13．解不等式组时要分别求出两个不等式的解集，然后在数轴上取交集。写成1≤x<