2019年高考真题和模拟题分项汇编数学(文)：专题09不等式、推理与证明(含解析).pdf

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1、2019 年高考真题和模拟题分项汇编数学（文）专题 09 不等式、推理与证明1【2019 年高考全国I 卷文数】古希腊时期，人们认为最美人体的头顶至肚脐的长度与肚脐至足底的长度之比是512（512 0.618，称为黄金分割比例)，著名的“断臂维纳斯”便是如此此外，最美人体的头顶至咽喉的长度与咽喉至肚脐的长度之比也是512若某人满足上述两个黄金分割比例，且腿长为105 cm，头顶至脖子下端的长度为26 cm，则其身高可能是A165 cm B175 cm C185 cm D 190 cm【答案】B【解析】方法一：如下图所示.依题意可知：5151,22ACABCDBC，腿长为 105 cm 得，即1

2、05CD，5164.892ACCD，64.89 105169.89ADACCD，所以 AD169.89.头顶至脖子下端长度为26 cm，即 AB26，第 1 页，共 18 页42.07512ABBC，=+68.07AC AB BC，110.15512ACCD，+68.07+110.15=178.22AC CD，所以178.22AD.综上，169.89b时，2abab当且仅当ab时取等号，则当4ab时，有24abab，解得4ab，充分性成立；当=1,=4ab时，满足4ab，但此时=54a+b，必要性不成立，综上所述，“4ab”是“4ab”的充分不必要条件.【名师点睛】易出现的错误有，一是基本不等

3、式掌握不熟，导致判断失误；二是不能灵活的应用“赋值法”，通过特取,a b的值，从假设情况下推出合理结果或矛盾结果.8【2019 年高考全国II 卷文数】若变量x，y 满足约束条件23603020 xyxyy，则 z=3x y 的最大值是 _.【答案】9【解析】画出不等式组表示的可行域，如图中阴影部分所示，第 6 页，共 18 页文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C

4、5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2

5、U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C

6、8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A1

7、0C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3

8、L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G

9、8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3

10、A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9阴影部分表示的三角形ABC 区域，根据直线30 xyz中的z表示纵截距的相反数，当直线3zxy过点3,0C（）时，z取最大值为 9【名师点睛】本题考查线性规划中最大值问题，渗透了直观想象、逻辑推理和数学运算素养采取图解法，利用数形结合思想解题搞不清楚线性目标函数的几何意义致误，从线性目标函数对应直线的截距观察可行域，平移直线进行判断取最大值还是最小值9【2019 年高考全国II 卷文数】中国有悠久的金石文化，印信是金石文化的代表之一印信的形状多为长方体、正方体或圆柱体，但南北朝时期的官员独孤信的印信形状是“半正多面体”（图 1）

11、半正多面体是由两种或两种以上的正多边形围成的多面体半正多面体体现了数学的对称美图 2 是一个棱数为48 的半正多面体，它的所有顶点都在同一个正方体的表面上，且此正方体的棱长为1则该半正多面体共有_个面，其棱长为_（本题第一空2 分，第二空3 分）【答案】26，21【解析】【答案】26，21【解析】由图可知第一层（包括上底面）与第三层（包括下底面）各有 9 个面，计 18 个面，第二层共有8 个面，所以该半正多面体共有18826个面如图，设该半正多面体的棱长为x，则ABBEx，延长CB与FE交于点G，延长BC交正方体棱于H，由半正多面体对称性可知，BGE为等腰直角三角形，第 7 页，共 18 页

12、文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5

13、H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U

14、1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8

15、A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10

16、C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L

17、2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8

18、C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A922,2(21)122BGGECHxGHxxx，12121x，即该半正多面体棱长为21【名师点睛】本题立意新颖，空间想象能力要求高，物体位置还原是关键，遇到新题别慌乱，题目其实很简单，稳中求胜是关键立体几何平面化，无论多难都不怕，强大空间想象能力，快速还原图形10

19、【2019 年高考北京卷文数】若 x，y 满足2,1,4310,xyxy则yx的最小值为 _，最大值为 _【答案】3；1【解析】根据题中所给约束条件作出可行域，如图中阴影部分所示.设zyx，则=+y x z，求出满足在可行域范围内z的最大值、最小值即可，即在可行域内，当直线=+y x z的纵截距最大时，z 有最大值，当直线=+y x z的纵截距最小时，z 有最小值.由图可知，当直线=+y x z过点 A 时,z 有最大值，联立24310 xxy，可得23xy，即(2,3)A，所以max321z；当直线=+y x z过点(2,1)B时，z 有最小值，所以min123z.第 8 页，共 18 页文

20、档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H

21、3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1

22、U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A

23、9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C

24、5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2

25、U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C

26、8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9【名师点睛】本题是简单线性规划问题的基本题型，根据“画、移、解”等步骤可得解.题目难度不大，注重了基础知识、基本技能的考查.11【2019 年高考天津卷文数】设0,0,24xyxy，则(1)(21)xyxy的最小值为 _.【答案】92【解析】(1)(21)22125

27、25xyxyyxxyxyxyxyxy.因为0,0,24xyxy，所以2422xyxy，即22,02xyxy，当且仅当22xy时取等号成立.又因为192255=22xy，所以(1)(21)xyxy的最小值为92.【名师点睛】使用基本不等式求最值时一定要验证等号是否能够成立.12【2019 年高考北京卷文数】李明自主创业，在网上经营一家水果店，销售的水果中有草莓、京白梨、西瓜、桃，价格依次为60 元/盒、65 元/盒、80 元/盒、90 元/盒为增加销量，李明对这四种水果进行促销：一次购买水果的总价达到120 元，顾客就少付x 元每笔订单顾客网上支付成功后，李明会得到支付款的80%当 x=10 时

28、，顾客一次购买草莓和西瓜各1 盒，需要支付 _元；在促销活动中，为保证李明每笔订单得到的金额均不低于促销前总价的七折，则x 的最大值为 _第 9 页，共 18 页文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H

29、3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1

30、U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A

31、9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C

32、5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2

33、U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C

34、8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9【答案】130；15.【解析】（1）10 x,顾客一次购买草莓和西瓜各一盒,需要支付608010130

35、元.（2）设顾客一次购买水果的促销前总价为y元,120y元时,李明得到的金额为80%y,符合要求.120y元时,有80%70%yxy恒成立,即87,8yyxy x,即min158yx元.所以x的最大值为 15.【名师点睛】本题主要考查不等式的概念与性质?数学的应用意识?数学式子变形与运算求解能力,以实际生活为背景，创设问题情境，考查学生身边的数学，考查学生的数学建模素养.13（四川省棠湖中学2019 届高三高考适应性考试数学（理)试题）已知集合(1)(4)0Axxx，2log2Bxx，则ABA2,4B1,C0,4D2,【答案】C【解析】(1)(4)01,4Ax xx，2log20,4Bxx，故

36、0,4AB，故选 C.【名师点睛】本题考查集合的交集，属于基础题，解题时注意对数不等式的等价转化.14【广东省韶关市2019 届高考模拟测试（4 月)数学试题】若x，y满足约束条件22201yxxyy，则zxy的最大值为A35-B12C5 D6【答案】C【解析】变量x，y满足约束条件的可行域如图中阴影部分所示：目标函数zxy是斜率等于1、纵截距为z的直线，第 10 页，共 18 页文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U

37、1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8

38、A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10

39、C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L

40、2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8

41、C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A

42、10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O

43、3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9当直线经过可行域的A点时，纵截距z取得最小值，则此时目标函数z取得最大值，由1220yxy可得(4,1)A，目标函数zxy的最大值为：5 故选：C【名师点睛】本题考查线性规划的简单应用，考查计算能力以及数形结合思想的应用15【山东省实验中学等四校2019 届高三联合考试理科数学试题】已知实数x，y满足约束条件202201xyxyx，则目标函数21yzx的最小值为A23B54C43D12【答案】B【解析】作出不等式组对应的平面区域如图：目标函数21yzx的几何意义为动点,Mx

44、y到定点1,2D的斜率，当M位于11,2A时，此时DA的斜率最小，此时min1252114z故选 B第 11 页，共 18 页文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX

45、9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码

46、:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4

47、 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8

48、ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档

49、编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3

50、O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9文档编码:CY3A10C5H3O4 HJ4O3L2U1U8 ZX9T1G8C8A9【名师点睛】本题主要考查线性规划的应用以及两点之间的斜率公式的计算，利用z 的几何意义，通过数形结合是解决本题的关键16【黑龙江省大庆市第