2025-2026学年北京市朝阳区九年级数学中考一模模拟试卷（含答案详解与评分标准）

2025-2026学年北京市朝阳区九年级数学中考一模模拟试卷（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分注意事项1．本试卷面向九年级中考一模考前检测，重在查验阶段复习效果和综合运用能力。2．答题前请将学校、班级、姓名和考号填写清楚。选择题按题号填写选项，解答题写出必要的文字说明、计算步骤或推理依据。3．全卷共三部分，选择题10题30分，填空题6题18分，解答题6题72分，合计120分。4．除作图题可使用铅笔外，其余内容建议使用黑色字迹笔书写；作答保持卷面整洁。一、选择题（本题共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．计算3²-2的值是（）。A．6B．7C．8D．92．若aᵐ·a²=a⁷，且a≠0，则m的值为（）。A．2B．3C．5D．63．下列点中，在一次函数y=2x-1的图象上的是（）。A．（1，1）B．（0，1）C．（-1，1）D．（2，2）4．不等式2x-1<5的解集是（）。A．x<2B．x<3C．x>2D．x>35．方程x²-4x+3=0的两个根是（）。A．-1，-3B．1，-3C．1，3D．-1，36．一组数据6，8，9，10，12的平均数与中位数分别是（）。A．9，9B．9，10C．8，9D．10，97．两条平行直线被第三条直线所截，若一组同旁内角中较小角为65°，则另一同旁内角为（）。A．65°B．105°C．110°D．115°8．半径为5的圆中，一条弦到圆心的距离为3，则这条弦的长为（）。A．6B．8C．10D．129．袋中有3个红球、2个白球，除颜色外完全相同。随机摸出1个球，摸到红球的概率是（）。A．1/5B．2/5C．3/5D．4/510．抛物线y=-(x-2)²+3的对称轴与最大值分别为（）。A．x=2，最大值3B．x=-2，最大值3C．x=2，最大值-3D．x=-2，最大值-3选择题答题区12345678910二、填空题（本题共6小题，每小题3分，共18分）11．用科学记数法表示0.000056，结果为__________。12．分解因式：x²-9=__________。13．若反比例函数y=k/x的图象经过点（2，-3），则k=__________。14．正六边形的每一个外角等于__________度。15．在△ABC中，点D在AB上，点E在AC上，DE∥BC。若AD=3，AB=6，DE=5，则BC=__________。16．数列1，3，5，7，…的第n项为aₙ=2n-1，则前12项的和为__________。填空题答题区111213141516三、解答题（本题共6小题，共72分）解答应写出文字说明、计算过程或推理依据。17．（12分）计算与解方程。（1）化简：(x²-9)/(x²+6x+9)÷(x-3)/(x+3)，其中x=6；（2）解方程组：{x+y=7，2x-y=8}。__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18．（12分）某校九年级为了做好中考一模后的阶段复习，购进甲、乙两类数学复习资料共70本，甲类每本18元，乙类每本25元，共花费1470元。（1）求甲、乙两类资料各购买多少本；（2）学校准备再增购20本同类资料，其中乙类不少于12本，且总费用不超过450元。求本次增购方案。______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19．（12分）在平面直角坐标系中，直线AB经过点A（-3，0）和点B（0，3）。点P（2，5）在该直线上，反比例函数y=k/x的图象也经过点P。（1）求直线AB的表达式；（2）求反比例函数的表达式，并判断点Q（-2，-5）是否在该反比例函数图象上；（3）求直线AB与反比例函数图象的两个交点坐标。__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20．（12分）在直角三角形ABC中，∠ACB=90°，AC=6，BC=8。点H为点C到斜边AB的垂足，即CH⊥AB。示意图如下，图中标注仅用于说明已知条件。（1）求AB的长；（2）求CH、AH、BH的长；（3）以C为圆心，CH为半径作圆，判断直线AB与该圆的位置关系，并说明理由。______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21．（12分）为了解一模后学生数学复习状况，某校随机抽取20名九年级学生的数学阶段检测成绩（单位：分）：68，72，75，78，80，82，82，85，86，88，89，90，90，91，92，94，95，96，98，100。（1）求这组数据的中位数和众数；（2）若该校九年级共有360名学生，估计成绩不低于90分的学生人数；（3）从不低于90分的9名学生中随机选2名参加经验交流，其中男生5名、女生4名。求恰好选到1名男生和1名女生的概率。______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22．（12分）已知抛物线y=-x²+4x+5与x轴交于A、B两点（点A在点B左侧），与y轴交于点C。点P在该抛物线第一象限部分，横坐标为t，且0<t<5。过点P作PQ⊥x轴于点Q。（1）求点A、B的坐标和抛物线顶点D的坐标；（2）用含t的式子表示△ABP的面积S，并求S的最大值；（3）当S=24时，求点P的坐标。____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析评分说明：选择题和填空题按最终结果给分；解答题按主要步骤、推理依据和结论分层给分。若解法不同但推理正确、结果一致，可参照对应步骤赋分。一、选择题答案与关键理由1．答案：B。解析：3²-2=9-2=7。2．答案：C。解析：aᵐ·a²=aᵐ⁺²=a⁷，因此m+2=7，m=5。3．答案：A。解析：把x=1代入y=2x-1，得y=1，点（1，1）满足表达式。4．答案：B。解析：2x-1<5，移项得2x<6，所以x<3。5．答案：C。解析：x²-4x+3=(x-1)(x-3)，故两根为1和3。6．答案：A。解析：平均数为(6+8+9+10+12)/5=9，有序数据中位数为9。7．答案：D。解析：两直线平行，同旁内角互补，另一角为180°-65°=115°。8．答案：B。解析：半径、弦心距和半弦构成直角三角形，半弦长为√(5²-3²)=4，弦长为8。9．答案：C。解析：共有5个球，其中红球3个，摸到红球的概率为3/5。10．答案：A。解析：y=-(x-2)²+3为顶点式，开口向下，对称轴为x=2，最大值为3。二、填空题答案与解析11．答案：5.6×10⁻⁵。解析：0.000056的小数点向右移动5位得到5.6，因此指数为-5。12．答案：(x+3)(x-3)。解析：x²-9=x²-3²，利用平方差公式得(x+3)(x-3)。13．答案：-6。解析：将点（2，-3）代入y=k/x，得-3=k/2，所以k=-6。14．答案：60。解析：正多边形外角和为360°，正六边形每个外角为360°÷6=60°。15．答案：10。解析：由DE∥BC可得△ADE∽△ABC，因此DE/BC=AD/AB=3/6=1/2，解得BC=10。16．答案：144。解析：前n个正奇数和为n²，当前12项和为12²=144。三、解答题答案详解与评分标准17．（12分）（1）原式=[(x-3)(x+3)]/(x+3)²×(x+3)/(x-3)。化简得原式=1。当x=6时，原式=1。原式有意义的条件为x≠3且x≠-3，题设取值x=6满足条件。（2）由x+y=7，得y=7-x。代入2x-y=8，得2x-(7-x)=8，即3x