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1、1/7 阅读理解型问题一、填空题1(2018 年 江 苏 南 通 三 模)如 图 所 示 的 程 序 是 函 数 型 的 数 值 转 换 程 序,其 中22x当输入的x 值为32时,输出的y值为.答案:12.三、解答题1、(2018 年浙江金华一模)阅读材料:如图,ABC 中,AB=AC,P 为底边 BC 上任意一点,点P 到 两 腰 的 距 离 分 别 为21,rr,腰 上 的 高 为h,连 结AP,则A B CA C PA B PSSS,即:hABrACrAB21212121,hrr21(1)理解 与应 用如果把“等腰三角形”改成“等边三角形”,那么P 的位置可以由“在底边上任一点”放宽为
2、“在三角形内任一点”,即:已知边长为2 的等边 ABC 内任意一点P 到各边的距离分别 为1r,2r,3r,试证明:1233rrr.(2)类比与推理边长为 2 的正方形内任意一点到各边的距离的和等于;(3)拓展 与延伸若边长为2 的正 n 边形 A1A2An 内部任意一点P 到各边的距离为nrrr,21,请问nrrr21是否为定值(用含n 的式子表示),如果是,请合理猜测出这个定值。h r2r1CABPr1r2r3hCPBA输入 x 的值2(11)yxx2(12)yxx2(21)yxx输出 y的值2/7 图1ACDB图2FOAECDB图3ACDB答案:(1)分别连接AP,BP,CP,由ABPB
3、CPACPABCSSSS可证得123rrrh,再求得等边三角形边的高为3,即可.(2)4.(3)00180tan(90)nn2、(2018 年北京市延庆县一诊考试)阅读下面材料:小红遇到这 样一个问题,如图1:在 ABC 中,ADBC,BD=4,DC=6,且 BAC=45,求线段 AD 的长.小红是这样想的:作ABC 的外接圆 O,如图2:利用同弧所对圆周角和圆心角的关系,可以知道BOC=90,然后过O 点作 OEBC 于 E,作 OFAD 于 F,在 RtBOC中可以求出 O 半径及 OE,在 Rt AOF 中可以求出AF,最后利用AD=AF+DF 得以解决此题。请你回答图2中线段 AD 的
4、长.参考小红思考问题的方法,解决下列问题:如图 3:在 ABC 中,AD BC,BD=4,DC=6,且 BAC=30,则线段 AD 的长.解:(1)12(2)3 11+5 3文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4
5、F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9
6、G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R
7、2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:C
8、Q4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 H
9、F9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE
10、2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C83/7 EABCD4、(2018 双柏县学业水平模拟考试)数学课上,李老师出示了如下框中的
11、题目在等边三角形 ABC中,点 E在AB上,点D在CB的延长线上,且ED=EC,如图.试确定线段 AE与DB的大小关系,并说明理由.EABCD小敏与同桌小聪讨论后,进行了如下解答:(1)特殊情况,探索结论当点 E 为 AB 的中点时,如图1,确定线段AE与 DB 的大小关系,请你直接写出结论:AEDB(填“”,“”,“0 时,163534bbb,得b=4,此时,坐标三角形面积为332;文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L
12、7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6
13、W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F
14、6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G
15、9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2
16、Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ
17、4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF
18、9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C86/7 当b0 时,163534bbb,得b=4,此时,坐标三角形面积为332.综上,当函数y43xb的坐标三角形周长为16 时,面积为332(12 分)8、(2018 山东省德州三模)在中央电视台第2 套购物街栏目中,有一个精彩刺激的游戏幸运大转盘,其规则如下:游戏工具是一个可绕轴心自由转动的圆形转盘,转盘按圆心角均匀划分为20 等分,并在其边缘标记5、10、15、100 共 20 个 5 的整数倍数,游戏时,选手可旋转转盘,待转盘停止时,指针所指的数即为本次游戏的得分
19、;每个选手在旋转一次转盘后可视得分情况选择是否再旋转转盘一次,若只旋转一次,则以该次得分为本轮游戏的得分,若旋转两次则以两次得分之和为本轮游戏的得分;若某选手游戏得分超过100 分,则称为“爆掉”,该选手本轮游戏裁定为“输”,在得分不超过 100 分的情况下,分数高者裁定为“赢”;遇到相同得分的情况,相同得分的选手重新游戏,直到分出输赢现有甲、乙两位选手进行游戏,请解答以下问题:(1)甲已旋转转盘一次,得分65 分,他选择再旋转一次,求他本轮游戏不被“爆掉”的概率(2)若甲一轮游戏最终得分为90 分,乙第一次旋转转盘得分为85 分,则乙还有可能赢吗?赢的概率是多少?(3)若甲、乙两人交替进行游
20、戏,现各旋转一次后甲得85 分,乙得65 分,你认为甲是否应选择旋转第二次?说明你的理由答案:解:(1)甲可取5、10、15、20、25、30、35,2分P(不爆掉)=2073 分(2)乙有可能赢,4 分乙可取 5、10、15,6 分P(乙赢)=2037 分(3)甲选择不转第二次.8 分理由是:甲选择不转第二次,乙必须选择旋转第二次,此时 P(乙赢)=203,乙获胜的可能性较小10 分文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L
21、7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6
22、W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F
23、6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G
24、9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2
25、Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ
26、4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF
27、9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C87/7 或“甲 若 选 择 转 第 二 次,P(甲 爆 掉)=2017,甲 输 而 乙 获 胜 的 可 能 性 较大”9、(2018 年浙江金华一模)(本题10 分)为了探索代数式221825xx的最小值,小明巧妙的运用了“数形结合”思想具体方法是这样的:如图,C 为线段 BD 上一动点,分别过点B、D 作,ABBD EDBD,连结AC、EC已知AB=1,DE=5,BD=8,设BC=x则21ACx,2825CEx则问题即转化成求AC+CE的最小值(1)我们知道当A、C、E
28、 在同一直线上时,AC+CE 的值最小,于是可求得221825xx的最小值等于,此时x;(2)请你根据上述的方法和结论,试构图求出代数式224129xx的最小值.答案:(1)10,43 (2)13.BDAEC312-x122BDAEFC文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5
29、HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 Z
30、E2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编
31、码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R
32、5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5
33、 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8文档编码:CQ4F6P10V2R5 HF9G9L7C4E5 ZE2R2Q6W7C8