九年级数学（七）参考答案_2026河北中考麒麟卷数学

2026 年初中毕业班（九年级）练习 数学（七） 参考答案 一、选择题（本大题共12小题，每小题3分，共36分） 1-5ACBAC 6-10BDACA 11-12CD 11.C 【解析】设桌面的宽为x，则2c＝4x，即c＝2x ax 5 由题意，  ，得a＝4x＝2c，故①正确 4x 4 又b＋x＝a 3 则b＝a－x＝3x＝ c，故②错误 2 故选：C 12.D 【解析】甲：设点P表示x 则P 表示的数为x＋2，P 表示的数为x＋2－4＝x－2 1 2 ∵P，P 表示的数互为相反数 1 2 ∴x＋2＋x－2＝0，解得：x＝0 ∴点P表示0，故甲说法正确 乙：∵点P表示-1 ∴P 表示的数为-1＋2＝1 1 P 表示的数为1－4＝-3 2 P 表示的数为-3＋6＝3 3 P 表示的数为3－8＝-5 4 P 表示的数为-5＋10＝5 5 P 表示的数为5－12＝-7 6 ∴当n为奇数时，P ＝n；当n为偶数时，P ＝（-1）n＋1（n＋1） n n ∵点P 到原点的距离为9 n ∴n＝9或n＝8，故乙说法错误 丙：设点P表示x ∴P 表示的数为x＋2 1 P 表示的数为x＋2－4＝x－2 2 P 表示的数为x－2＋6＝x＋4 3 P 表示的数为x＋4－8＝x－4 4 P 表示的数为x－4＋10＝x＋6 5 P 表示的数为x＋6－12＝x－6 6 ∴当n为奇数时，P ＝x＋n＋1；当n为偶数时，P ＝x－n n n ∴|P －P |＝|x＋n＋1－（x－n＋1）|＝2n，故丙说法错误 n n－1 综上可知：甲对，乙、丙不对 故选：D 二、填空题（本大题共4小题，每小题3分，共12分） 13.2（答案不唯一） 数学练习（七） 参考答案 第 1 页 共 7 页14.8 3 15.5 【解析】第1个化学式中有1个C和4个H 第2个化学式中有2个C和6个H 第3个化学式中有3个C和8个H …… 以此类推，第n个化学式中有n个C和（2n＋2）个H ∵烷烃C H 化学式中的m，n满足此规律 n m ∴m＝2n＋2 ∵12n＝5m ∴12n＝5（2n＋2） ∴n＝5 故答案为：5 16. 4 2 2 【解析】如下图 ∵点C为⊙B上一点，BC＝4 在x轴上取OD＝OA＝8，连接BD，交⊙B于点E ∵AM＝CM，OD＝OA ∴OM是△ACD的中位线 1 ∴OM＝ CD 2 当OM最小时，即CD最小 而D，B，C三点共线时，即当C在BD与⊙B的交点处时，OM最小，此时点E与点C重合 ∵OB＝OD＝8，∠BOD＝90° ∴BD＝8 2 ∴CD＝8 2－4 1 ∴OM CD＝4 2 2 2 即OM的最小值为4 2 2 故答案为：4 2 2 三、解答题（本大题共8小题，共72分. 解答应写出文字说明、证明过程或演算步骤） 17. 解：（1）x＋2≥1 x≥1－2 x≥-1·································································································1分 在数轴上表示如下图·············································································2分 数学练习（七） 参考答案 第 2 页 共 7 页（2）10－3x＞1 -3x＞1－10 -3x＞-9 x＜3·············································································································4分 此解集在数轴上表示如上图所示········································································5分 （3）-1，0，1，2··································································································7分 18. （1）二，一····································································································2分 x2 1 （x1）2 （2）解：习题1：原式＝  x1 x1 x2 1 x2 2x1 ＝  ·······························································5分 x1 x1 2x ＝- ···············································································8分 x1 x2 习题2： －x＝1 x1 x2－x（x＋1）＝x＋1·············································································3分 x2－x2－x＝x＋1 1 x＝- ············································································· 5分 2 1 检验：当x＝- 时，x＋1≠0······································································7分 2 1 ∴x＝- 是原方程的解··············································································8分 2 19. （1）证明：∵四边形ABCD是菱形 ∴BC∥AD ∴∠EAD＝∠EFC，∠EDA＝∠ECF ∵E是菱形ABCD的边CD的中点 ∴DE＝CE ∴△ADE≌△FCE（AAS）·································································4分 （2）解：∵E为菱形ABCD的边CD的中点 ∴AD＝CD＝2DE＝4 ∵∠AED＝90° ∴由勾股定理可得EA＝2 3 ······································································6分 ∵△ADE≌△FCE ∴EF＝EA ∴AF＝2EA＝4 3 ····················································································8分 20. 解：（1）如下图，直线l即为所求 ·······································································3分 数学练习（七） 参考答案 第 3 页 共 7 页（2）由（1）可知，EF为OC的垂直平分线 ∴CG＝OG，∠EFO＝90° ∵OC＝OA＝6 ∴OF＝CF＝3 由题意可知∠DOC＝30° OF 3 ∴cos∠COG＝  OG 2 ∴OG＝2 3 ∴CG＝2 3··································································································6分 （3）由题意可知∠AOC＝∠DOC＝30° 1 ∴∠ABD＝ ∠AOD＝30° 2 ∴OC∥BD ∴△FOG∽△EDG OG OF ∴  DG DE ∵OD＝6，OG＝2 3 ∴DG＝6-2 3 ∴DE＝3 33······························································································8分 21. 解：（1）由图可知，A（优秀，分数范围x≥26）有4名，该公司共有员工4＋7＋6＋3＝20（名） 4 ∴m＝ ×100%＝20%·········································································3分 20 （2）该公司共有20名员工，则成绩中位数23是第10名与第11名员工成绩的平均数 ∵该公司成绩排名（从高到低）第10名员工的成绩为24分 设排名为第11名员工的成绩为n分 24n 则 ＝23，解得n＝22···············································································6分 2 答：排名为第11名员工的成绩为22分 （3）∵20＜21 ∴员工进修情况要发生变化，即成绩平均分需不低于21分，则总分需要增加20分 ∵该公司部分员工有科研技术奖励分值，且每项科研技术奖励分值为10分 ∴员工至少有2项科研技术··············································································9分 22. （1）解：设直线MN的解析式为：y＝kx＋b ∵每个台阶宽、高均分别为30cm和20cm ∴M（0，160），N（240，0） b160 将M（0，160）和N（240，0）代入解析式得： 0240kb  2 k  解得： 3  b160 2 ∴y＝- x＋160··················································································3分 3 数学练习（七） 参考答案 第 4 页 共 7 页由题意得，点B 的坐标为（210，20） 1 2 当x＝210时，y＝- ×210＋160＝20 3 ∴点B（210，20）在直线MN上···························································4分 1 （2）在··············································································································· 5分 2 y＝- x＋180································································································7分 3 （3）解：把N（240，0）代入y＝mx－260m＋180（m≠0） 得240m－260m＋180＝0 解得m＝9 把M（0，160）代入y＝mx－260m＋180（m≠0） 得-260m＋180＝160 1 解得m＝ 13 1 ∴ ≤m≤9···························································································9分 13 23. 解：（1）依题意，设嘉嘉第一个蛙跳的路线抛物线L 的解析式为y＝a（x－1）2＋0.4 1 代入（0，0），得0＝a＋0.4 解得：a＝-0.4 ∴嘉嘉第一个蛙跳的路线抛物线L 的解析式为y＝-0.4（x－1）2＋0.4（或y＝-0.4x2＋0.8x） 1 ········································································································2分 （2）①∵以第一个蛙跳起跳点为原点，落在点A处，对称轴为x＝1 ∴A（2，0） ∵第二个蛙跳路线为抛物线L：y＝a（x－h）2＋k（a≠0），其开口大小和方向均与第一个蛙 2 18 跳的路线抛物线L 相同，第二个蛙跳路线最高点为 m 1 125 18 ∴L：y＝-0.4（x－h）2＋ 2 125 又L 过点A（2，0） 2 18 ∴-0.4（2－h）2＝- 125 解得：h＝1.4（舍去），h＝2.6 1 2 第二个蛙跳落地点与点A的距离为：（2.6－2）×2＝1.2（m）······························5分 1.2＋2＝3.2（m） ∴第二个蛙跳落地点距离第一个蛙跳的起跳点的距离为3.2m································6分 ②嘉嘉在第二个蛙跳中不会越过可调节支撑杆······················································7分 理由如下： 18 由①可知抛物线L 的解析式为y＝-0.4（x－2.6）2＋ 2 125 18 当x＝3时，y＝-0.4×（3－2.6）2＋ ＝0.08 125 ∵0.08＜0.12 ∴嘉嘉在第二个蛙跳中不会越过可调节支撑杆···················································9分 数学练习（七） 参考答案 第 5 页 共 7 页9 （3）2≤h≤ ·····································································································11分 4 【解析】∵点P是抛物线L 与直线y＝mx的交点 1 y0.4(x1)2 0.4 ∴联立 ymx x22.5m x0 解得 或 y2m2.5m2 y0 ∴点P的坐标为（2－2.5m，2m－2.5m2） 由题意可知L 的解析式为y＝-0.4（x－h）2＋0.4 2 代入P（2－2.5m，2m－2.5m2），得-0.4（2－2.5m－h）2＋0.4＝2m－2.5m2 整理得（2－2.5m－h）2＝6.25m2-5m+1 （2－2.5m－h）2＝（2.5m－1）2 解得h＝1（舍去）或h＝3－5m 3 1 ∵ ≤m≤ 20 5 9 ∴2≤h≤ 4 3 1 9 ∴当 ≤m≤ ，且抛物线L 与抛物线L 的顶点的纵坐标恰好相等时，2≤h≤ 20 5 2 1 4 24. （1）①证明：∵四边形ABCD是正方形 ∴AB＝AD，∠BAD＝90° ∵△AEF是含有45°的直角三角尺 ∴△AEF是等腰直角三角形 ∴AE＝AF，∠EAF＝90° ∴∠BAD＝∠EAF ∴∠BAD－∠DAE＝∠EAF－∠DAE 即∠BAE＝∠DAF ∴△ABE≌△ADF（SAS） ∴∠ABG＝∠ADF·········································································3分 ②BE＝DF，BE⊥DF··················································································4分 证明：由①可知△ABE≌△ADF ∴BE＝DF ∵∠ABE＝∠ADF，∠AMB＝∠DMG ∴∠DGM＝∠BAM＝90° 即BE⊥DF ∴BE＝DF，BE⊥DF······································································7分 （2）解：∵△BAD是直角三角形，O是BD的中点 1 ∴OA＝ BD 2 由（1）知∠DGB＝90° ∴△BGD是直角三角形 1 ∴OG＝ BD 2 数学练习（七） 参考答案 第 6 页 共 7 页∴OA＝OG····························································································10分 （3）点G经过路线的长度为2 2π········································································12分 【解析】由（2）知，OA＝OD＝OG ∴点G的运动轨迹是以O为圆心，OA为半径的弧 ∵旋转角α从0°变化到60 ⌒ ∴此时点G的运动路线就是AG 取AB中点H，连接EH ∵∠BAE＝60°，AB＝12 1 AE＝ AB＝6 2 1 ∴AE＝AH＝ AB＝BH 2 ∴△AEH为等边三角形 ∴EH＝AH＝BH，∠AHE＝∠AEH＝60° ∴∠HBE＝∠BEH ∵∠AHE＝∠HBE＋∠BEH 1 ∴∠HBE＝∠BEH＝ ∠AHE＝30° 2 即∠ABE＝30° ∴∠OBG＝45°－30°＝15° 1 ∵OB＝OG＝ BD 2 ∴∠DOG＝30° ∵AB＝AD，O为BD中点 ∴∠AOB＝90° ∴∠AOG＝180°－∠AOB－∠DOG＝60° ∵AB＝12，BD＝ 2AB＝12 2 ∴OA＝OG＝6 2 ⌒ 60π6 2 ∴AG的长度为 ＝2 2π 180 即点G经过路线的长度为2 2π 数学练习（七） 参考答案 第 7 页 共 7 页