(完整word版)高中数学必修一和必修二第一二章综合试题(人教A版含答案)(word文档良心出品).pdf

第 1 页 共 8 页高一数学第二次月考模拟试题（必修一+二第一二章）时间：120 分钟分值：150 分一、选择题(每小题 5 分，共 60 分)1设集合A4,5,7,9，B3,4,7,8,9，全集UAB，则集合?U(AB)中的元素共有()A3个 B4 个 C 5个 D6 个2下列函数为奇函数的是()Ayx2 Byx3 Cy2x Dylog2x3函数y1xlog2(x3)的定义域是()AR B(3，)C(，3)D(3,0)(0，)4梯形1111A B C D（如图）是一水平放置的平面图形ABCD的直观图（斜二测），若11A D/y轴，11A B/x轴，1111223A BC D，111A D，则平面图形ABCD的面积是（）A.5 B.10 C.5 2 D.1025已知圆锥的表面积是底面积的3 倍，那么该圆锥的侧面展开图扇形的圆心角为（）A.120 B.150 C.180 D.2406已知f(x31)x1，则f(7)的值，为()A.371 B.37 1 C3 D2 7已知 log23a，log25b，则 log295等于()Aa2b B2ab C.a2b D.2ab8函数yx2x(1x3)的值域是()A0,12 B 14，12 C 12，12 D34，12 9下列四个图象中，表示函数f(x)x1x的图象的是()A1 B1 C1 D1 O1 第 2 页 共 8 页10函数yx28x16 在区间 3,5上()A没有零点 B有一个零点 C有两个零点 D有无数个零点11给出以下四个命题：如果一条直线和一个平面平行，经过这条直线的一个平面和这个平面相交，那么这条直线和交线平行；如果一条直线和一个平面内的两条相交直线都垂直，那么这条直线垂直于这个平面；如果两条直线都平行于一个平面，那么这两条直线互相平行；如果一个平面经过另一个平面的一条垂线，那么些两个平面互相垂直.其中真命题的个数是()A4 B3 C 2 D 1 12已知f(x)是定义在(0，)上的增函数，若f(x)f(2 x)，则x的取值范围是()Ax1 Bx1 C0 x2 D1x2 二、填空题(每小题 5 分，共 20 分)13已知集合A x|x 1或 2x3，Bx|2x4，则AB _.14函数ylog234x的定义域为 _15据有关资料统计，通过环境整治，某湖泊污染区域S(km2)与时间t(年)可近似看作指数函数关系，已知近两年污染区域由0.16 km2降至 0.04 km2，则污染区域降至0.01 km2还需要 _年16空间四边形ABCD中，P、R分别是AB、CD的中点，PR=3、AC=4、BD=2 5,那么AC与BD所成角的度数是_.三、解答题(写出必要的计算步骤，只写最后结果不得分，共70 分)17(10 分)已知集合Ax|1 x4，Bx|xa0，解得x 3 且x0.答案：D 4 解析：梯形1111A B C D上底长为2，下底长为3 腰梯形11A D长为 1，腰11A D与下底11C D的夹角为45，所以梯形1111A B C D的高为22，所以梯形1111A B C D的面积为125 2+=224（2 3），根据2S=S4直观平面可知，平面图形ABCD的面积为5.答案：A 5解 析：由22rr3 rl知 道2lr所 以 圆 锥 的 侧 面 展 开 图 扇 形 圆 心 角 度 数 为13603601802rl，故选 C答案：C 6解析：令x317，得x2，f(7)3.答案：C 7解析：log295log29 log252log23log252ab.答案：B 8解析：画出函数yx2x(1x3)的图象，由图象得值域是 14，12 答案：B 9解析：函数yx，y1x在(0，)上为增函数，所以函数f(x)x1x在(0，)上为增函数，故满足条件的图象为A.答案：A 10解析：yx28x16(x4)2，函数在 3,5上只有一个零点4.答案：B 11解析：因为正确，故选B12解析：由题目的条件可得x02x0 x2x，解得 1x2，故答案应为D.答案：D 二、填空题(每小题 5 分，共 20 分)13答案：x|x4 文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9第 6 页 共 8 页14解析：根据对数函数的性质可得log2(34x)0 log21，解得 3 4x1，得x12，所以定义域为(，12 答案：(，12 15解析：设Sat，则由题意可得a214，从而a12，于是S(12)t，设从 0.04 km2降至 0.01 km2还需要t年，则(12)t14，即t 2.答案：2 16、解析：如图，取AD中点Q，连PQ，RQ，则5PQ，2RQ，而PR=3，所以222PQRQPR，所以PQR为直角三角形，90PQR，即PQ与RQ成90的角，所以AC与BD所成角的度数是90.答案：90三、解答题(写出必要的计算步骤，只写最后结果不得分，共70 分)17(10 分)已知集合Ax|1 x4，Bx|xa0，(1)当a3 时，求AB；(2)若A?B，求实数a的取值范围解：(1)当a3 时，Bx|x30 x|x3，则有ABx|1 x3(2)Bx|xa0 x|xa，当A?B时，有a4，即实数a的取值范围是 4，)18(12 分)(1)计算：(279)12(lg5)0(2764)-13；(2)解方程：log3(6x9)3.解：(1)原式(259)12(lg5)0(34)313531434.(2)由方程 log3(6x9)3 得 6x93327，6x 3662，x2.经检验，x2 是原方程的解19(12 分)判断函数f(x)1ax1x312的奇偶性解：由ax10，得x0，文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9第 7 页 共 8 页函数定义域为(，0)(0，)，f(x)1ax1(x)312ax1axx312ax111axx3121ax1x312f(x)f(x)为奇函数20(12 分)如图，在长方体ABCDA1B1C1D1中，AB 2，BB1BC1，E为D1C1的中点，连结ED，EC，EB和DB(1)求证：平面EDB平面EBC；(2)求二面角EDBC的正切值.证明：(1)在长方体ABCDA1B1C1D1中，AB2，BB1BC1，E为D1C1的中点DD1E为等腰直角三角形，D1ED45同理C1EC4590DEC，即DEEC在长方体ABCD1111DCBA中，BC平面11DCCD，又DE平面11DCCD，BCDE又CBCEC，DE平面EBC平面DEB过DE，平面DEB平面EBC(2)解：如图，过E在平面11DCCD中作EODC于O在长方体ABCD1111DCBA中，面ABCD 面11DCCD，EO面ABCD过O在平面 DBC中作OFDB于F，连结EF，EFBDEFO为二面角EDBC的平面角利用平面几何知识可得OF51，(第20 题)又OE1，所以，tanEFO5 21.(12分)已知正方体1111ABCDA BC D，O是底ABCD对角线的交点.求证：（）OC1面11AB D；（2）1AC面11AB D证明：（1）连结11A C，设11111ACB DO连结1AO，1111ABCDA B C D是正方体11A ACC是平行四边形D1ODBAC1B1A1C文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9第 8 页 共 8 页11A CAC且11ACAC又1,O O分别是11,ACAC的中点，11O CAO且11O CAO11AOC O是平行四边形111,C OAOAO面11AB D，1C O面11AB D1C O面11AB D（2）1CC面1111A B C D11!CCB D又1111ACB D，1111B DAC C面111ACB D即同理可证11ACAB，又1111D BABB1AC面11AB D22(12 分)已知函数f(x)是正比例函数，函数g(x)是反比例函数，且f(1)1，g(1)1，(1)求f(x)，g(x)；(2)判断函数h(x)f(x)g(x)的奇偶性；(3)证明函数S(x)xf(x)g(12)在(0，)上是增函数解：(1)设f(x)k1x(k10)，g(x)k2x(k20)f(1)1，g(1)1，k1 1，k21.f(x)x，g(x)1x.(2)由(1)得h(x)x1x，则函数h(x)的定义域是(，0)(0，)，h(x)x1x(x1x)h(x)，函数h(x)f(x)g(x)是奇函数(3)证明：由(1)得S(x)x22.设x1，x2(0，)，且x1x2，则S(x1)S(x2)(x212)(x222)x21x22(x1x2)(x1x2)x1，x2(0，)，且x1x2，x1x20.S(x1)S(x2)0.S(x1)S(x2)函数S(x)xf(x)g(12)在(0，)上是增函数文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9文档编码:CK7L10M3Y9E2 HP10X4F2O1O3 ZJ6S9J7W8D9