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2019年新课标全国卷2理科数学模拟卷一.pdf

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1、考试大纲答案详解1 备考 2019 YIXIMATH 2019 年新课标全国卷2 理科数学模拟卷一(考试时间:120 分钟试卷总分值:150 分)第卷选择题(共 60分)一、选择题(本大题共 12小题,每题 5分,在每题给出的四个选项中,只有一项是符合题目要求的)1.已知 i 是虚数单位,则复数2+3i3-2i=()A.-2+i B.i C.2-i D.-i 2.已知集合 M=x|x2-4x2”是“x1”的充分不必要条件.则以下命题是真命题的是()A.pqB.(p)(q)C.(p)qD.p(q)5.已知点 A是抛物线 C1:y2=2px(p0)与双曲线 C2:?2?2-?2?2=1(a0,b0

2、)的一条渐近线的交点,假设点 A到抛物线 C1的焦点的距离为p,则双曲线 C2的离心率等于()A.2B.3C.5D.66.(?-1?)12的展开式中含 x 的正整数指数幂的项的个数是()A.1 B.2 C.3 D.4 7.假设数列 an是等差数列,则以下结论正确的选项是()A.假设 a2+a50,则 a1+a20 B.假设 a1+a30,则 a1+a20 C.假设 0a1?2?4D.假设 a10 8.考试大纲答案详解2 备考 2019 YIXIMATH 如图,正四棱锥 P-ABCD 底面的四个顶点 A,B,C,D 在球 O 的同一个大圆上,点 P 在球面上,假设 V正四棱锥P-ABCD=163

3、,则球 O 的外表积是()A.4B.8C.12D.169.已知变量 x,y满足线性约束条件 3?+?-2 0,?-?2,?-?-1,假设目标函数 z=kx-y仅在点(0,2)处取得最小值,则 k的取值范围是()A.k1 C.-1k 1 D.-3k0),且g(e)=a,e为自然对数的底数.(1)已知 h(x)=e1-xf(x),求曲线 h(x)在点(1,h(1)处的切线方程;(2)假设存在 x1,e,使得 g(x)-x2+(a+2)x成立,求 a 的取值范围;(3)设函数 F(x)=?(?),?1,?(?),?1,O 为坐标原点,假设对于 y=F(x)在 x-1时的图象上的任一点 P,在曲线 y

4、=F(x)(xR)上总存在一点 Q,使得?0),过点 P(-4,-2)的直线 l 的参数方程为?=-4+22?,?=-2+22?(t 为参数),直线l 与曲线 C分别交于点 M,N.(1)写出 C 的直角坐标方程和l 的普通方程;(2)假设|PM|,|MN|,|PN|成等比数列,求 a 的值.23.(本小题总分值 10分)选修 45:不等式选讲已知函数 f(x)=|x-1|+|x+1|.(1)求不等式 f(x)3的解集;(2)假设关于 x的不等式 f(x)a2-x2+2x在 R 上恒成立,求实数 a 的取值范围.参考答案1.B解析(方法一)2+3i3-2i=(2+3i)(3+2i)(3-2i

5、)(3+2i)=13i13=i.(方法二)2+3i3-2i=(2+3i)i(3-2i)i=(2+3i)i2+3i=i.2.A解析 M=x|0 x0,则 a1+a2=(a2-d)+(a5-3d)=(a2+a5)-4d.由于 d的正负不确定,因而 a1+a2的符号不确定,故选项 A 错误.假设 a1+a30,则 a1+a2=(a1+a3)-d.由于 d的正负不确定,因而 a1+a2的符号不确定,故选项 B 错误.假设 0a10.所以 a30,a40.所以?32-a2a4=(a1+2d)2-(a1+d)(a1+3d)=d20.所以 a3?2?4.故选项 C正确.由于(a2-a1)(a4-a2)=d(

6、2d)=2d2,而 d有可能等于 0,故选项 D 错误.8.D解析连接 PO,由题意知,PO底面 ABCD,PO=R,S正方形ABCD=2R2.因为 V正四棱锥P-ABCD=163,所以13 2R2 R=163,解得 R=2.所以球 O 的外表积是 16.9.D解析如图,作出题中不等式组所表示的平面区域.由 z=kx-y得 y=kx-z,要使目标函数 z=kx-y 仅在点 A(0,2)处取得最小值,则阴影部分区域在直线y=kx+2的下方,故目标函数线的斜率 k 满足-3k 1.10.D解析由该几何体的三视图可得其直观图为如下图的三棱锥,且从点 A 出发的三条棱两两垂直,AB=1,PC=6

7、,PB=a,BC=b.文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1

8、ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1

9、 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E

10、1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8

11、E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R

12、8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7

13、R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10考试大纲答案详解8 备考 2019 YIXIMAT

14、H 可知 PA2+AC2=a2-1+b2-1=6,即 a2+b2=8.故(a+b)2=8+2ab8+2(?+?2)2,即 a+b4,当且仅当 a=b=2 时,a+b 取得最大值,此时 PA=3,AC=3.所以该几何体的体积V=1312 1 3 3=12.11.C解析由?=2 3,BAC=30,可得 SABC=1,即 x+y+z=1.故1?+4?+9?=(1?+4?+9?)(x+y+z)=1+4+9+?+?+4?+4?+9?+9?14+4+6+12=36,当且仅当 x=16,y=13,z=12时等号成立.因此,f(x,y,z)的最小值为 36.12.D解析假设对于函数图象上的任意一点M(x1

15、,y1),在其图象上都存在点N(x2,y2),使OMON,则函数图象上的点的集合为“商高线”.对于,假设取 M(1,1),则不存在这样的点;对于,假设取 M(1,0),则不存在这样的点.都符合.故选 D.13.0解析假设输入 x=0.1,则 m=lg 0.1=-1.因为 m0),g(x)=aln x+c(c 为常数).g(e)=aln e+c=a+c=a.c=0.g(x)=aln x.由 g(x)-x2+(a+2)x,得(x-ln x)ax2-2x.当 x1,e时,ln x1x,且等号不能同时成立,ln x0.a?2-2?-ln?.a(?2-2?-ln?)max.文档编码:CD9K10A4A

16、5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4

17、A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A

18、4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10

19、A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K1

20、0A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K

21、10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9

22、K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10考试大纲答案详解12 备考 2019 YIXIMATH 设 t(x)=?2-2?-ln?,x1,e,

23、则 t(x)=(?-1)(?+2-2ln?)(?-ln?)2.x1,e,x-10,ln x1,x+2-2ln x0.t(x)0.t(x)在1,e上为增函数.t(x)max=t(e)=e2-2ee-1.ae2-2ee-1.(3)设 P(t,F(t)为 y=F(x)在 x-1时的图象上的任意一点,则 t-1.PQ的中点在 y轴上,点 Q 的坐标为(-t,F(-t).t-1,-t1.P(t,-t3+t2),Q(-t,aln(-t).?=-t2-at2(t-1)ln(-t)0,a(1-t)ln(-t)1.当 t=-1 时,a(1-t)ln(-t)1 恒成立,此时 aR.当 t-1 时,a1(1-?)l

24、n(-?),令 (t)=1(1-?)ln(-?)(t-1),则 (t)=(?-1)+?ln(-?)?(1-?)ln(-?)2.t-1,t-10,tln(-t)0.(t)=1(1-?)ln(-?)在(-,-1)内为增函数.当 t-时,(t)=1(1-?)ln(-?)0,(t)0.a0.综上,可知 a的取值范围是(-,0.22.解(1)曲线 C 的直角坐标方程为x2=2ay(a0),直线 l 的普通方程为 x-y+2=0.(2)将直线 l 的参数方程与 C 的直角坐标方程联立,得 t2-2 2(4+a)t+8(4+a)=0.(*)由 =8a(4+a)0,可设点 M,N 对应的参数分别为t1,t2,

25、且 t1,t2是方程(*)的根,则|PM|=|t1|,|PN|=|t2|,|MN|=|t1-t2|.由题设得(t1-t2)2=|t1t2|,即(t1+t2)2-4t1t2=|t1t2|.由(*)得 t1+t2=22(4+a),t1t2=8(4+a)0,则有(4+a)2-5(4+a)=0,解得 a=1 或 a=-4.文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4

26、A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A

27、4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10

28、A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K1

29、0A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K

30、10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9

31、K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD

32、9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10考试大纲答案详解13 备考 2019 YIXIMATH 因为 a0,所以 a=1.23.解(1)原不等式等价于?1,2?3.解得 x-32或 x32.故原不等式的解集为?|?-32或?32.(2)令 g(x)=|x-1|+|x+1|+x2-2x,则 g(x)=?2-4?,?1.当 x(-,1时,g(x)单调递减;当 x1,+)时,g(x)单调递增.故当

33、x=1 时,g(x)取得最小值 1.因为不等式 f(x)a2-x2+2x在 R 上恒成立,所以 a21,解得-1a 1.所以实数 a的取值范围是(-1,1).文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:C

34、D9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:

35、CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码

36、:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编

37、码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档

38、编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文

39、档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10文档编码:CD9K10A4A5C5 HT10A5Q7R8E1 ZZ2J4H3C7X10

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