2020年山西省中考市试卷及答案

2020年山西省中考市试卷及答案 数 学 一、选择题〔每题2分，共20分〕 1．比较大小：2 3〔填〝＞〞、〝=〞或〝＜〝〕． 2．山西有着丰富的旅行资源，如五台山、平遥古城、乔家大院等闻名景点，吸引了众多的海内外游客，2018年全省旅行总收入739.3亿元，那个数据用科学记数法可表示为 ． A 3．请你写出一个有一根为1的一元二次方程： ． B 4．运算：123= ． 1 C 5．如下图，A、B、C、D是圆上的点，170° ，A40°，D 那么C 度． 〔第5题〕 6．李师傅随机抽查了本单位今年四月份里6天的日用水量〔单位：吨〕结果如下：7，8，8，7，6，6，依照这些数据，估量四月份本单位用水总量为 吨． 7．如图，△ABC与△ABC是位似图形，且顶点都在格点上，那么位似中心的坐标是 ． y 11 A 10 9 8 7 6 5 A A  BD C 4 3 2 B C E O 1 B C O 1 2 3 4 5 6 7 8 9 10 11 12 x 〔第8题〕 〔第7题〕 8．如图，ABCD的对角线AC、BD相交于点O，点E是CD的中点，△ABD的周长为16cm，那么△DOE的周长是 cm． 3，那么当x1时，y的取值范畴是 ． x10．以下图案是晋商大院窗格的一部分，其中〝○〞代表窗纸上所贴的剪纸，那么第n个图9．假设反比例函数的表达式为y中所贴剪纸〝○〞的个数为 ． 〔2〕 〔1〕 〔第10题〕 …… 〔3〕 …… 二、选择题〔在以下各小题中，均给出四个备选答案，其中只有一个正确答案，请将正确答案的字母号填入下表相应的空格内，每题3分，共24分〕 11．以下运算正确的选项是〔 〕 A．aaa B．2623C．3x2·2x36x6 12．反比例函数y k的图象通过点2，3，那么k的值是〔 〕 x32A． B． C．6 D．6 232 0D．π31 1x2≥113．不等式组的解集在数轴上可表示为〔 〕 3x18 A． B． 0 1 2 3 4 0 C． D． 0 1 2 3 4 0 1 2 3 4 1 2 3 4 14．解分式方程1x1，可知方程〔 〕 2x22xA．解为x2 B．解为x4 C．解为x3 D．无解 15．如图是由几个相同的小正方体搭成的几何体的三视图，那么搭成那个几何体的小正方体的个数是〔 〕 主视图 左视图 俯视图 〔第15题〕 A．5 B．6 C．7 D．8 16．如图，AB是⊙O的直径，AD是⊙O的切线，点C在⊙O上，BC∥OD，AB2，OD3,那么BC的长为〔 〕 A． 2 3B．3 2C．m 3 2D．2 2CBO n n n 〔1〕 〔2〕 〔第17题〕 〔第16题〕 17．如图〔1〕，把一个长为m、宽为n的长方形〔mn〕沿虚线剪开，拼接成图〔2〕，D A成为在一角去掉一个小正方形后的一个大正方形，那么去掉的小正方形的边长为〔 〕 A．mn 2B．mn C．m 2 D．n 218．如图，在Rt△ABC中，ACB90°，BC3，AC4，AB的垂 直平分线DE交BC的延长线于点E，那么CE的长为〔 〕 A D E 3725A． B． C． 266三、解答题〔此题共76分〕 19．〔每题4分，共12分〕 〔1〕运算：x3x1x2 2 D．2 B C 〔第18题〕 x22x2〔2〕化简：2 x4x2 〔3〕解方程：x2x30 20．〔此题6分〕每个网格中小正方形的边长差不多上1，图1 中的阴影图案是由三段以格点为圆心，半径分不为1和2的圆弧围成． 〔1〕填空：图1中阴影部分的面积是 〔结果保留π〕； 〔2〕请你在图2中以图1为差不多图案，借助轴对称、平移或旋转设计 〔第20题 图1〕 一个完整的花边图案〔要求至少含有两种图形变换〕． 〔第20题 图2〕 21．〔此题8分〕依照山西省统计信息网公布的数据，绘制了山西省2004~2018固定电话和移动电话年末用户条形统计图如下： 万户 固定电话年末用户 1800 1689.5 移动电话年末用户 1600 1420.4 1400 1200 989.6 906.2 1000 897.8 885.4 859.0 803.0 753.8 800 721.3 600 400 200 0 2004 2005 2006 2007 2018 年份 〔第21题〕 〔1〕填空：2004~2018移动电话年末用户的极差是 万户，固定电话年末用户的中位数是 万户； 〔2〕你还能从图中猎取哪些信息？请写出两条． 222．〔此题8分〕某商场为了吸引顾客，设计了一种促销活动：在一个不透亮的箱子里放有4个相同的小球，球上分不标有〝0元〞、〝10元〞、〝20元〞和〝30元〞的字样．规定：顾客在本商场同一日内，每消费满200元，就能够在箱子里先后摸出两个球〔第一次摸出后不放回〕．商场依照两小球所标金额的和返还相应价格的购物券，能够重新在本商场消费．某顾客刚好消费200元． 〔1〕该顾客至少可得到 元购物券，至多可得到 元购物券； 〔2〕请你用画树状图或列表的方法，求出该顾客所获得购物券的金额不低于30元的概率． 23．〔此题8分〕有一水库大坝的横截面是梯形ABCD，AD∥BC，EF为水库的水面，点E在DC上，某课题小组在老师的带领下想测量水的深度，他们测得背水坡AB的长为12米，迎水坡上DE的长为2米，BAD135°求水深．〔精确到，ADC120°，，31.73〕 0.1米，21.41F 水深 B C 〔第23题〕 24．〔此题8分〕某批发市场批发甲、乙两种水果，依照以往体会和市场行情，估量夏季某一段时刻内，甲种水果的销售利润y甲〔万元〕与进货量x〔吨〕近似满足函数关系A D E y甲0.3x；乙种水果的销售利润y乙〔万元〕与进货量x〔吨〕近似满足函数关系y乙ax2bx〔其中a0，a，b为常数〕，且进货量x为1吨时，销售利润y乙为1.4万元；进货量x为2吨时，销售利润y乙为2.6万元． 〔1〕求y乙〔万元〕与x〔吨〕之间的函数关系式． 〔2〕假如市场预备进甲、乙两种水果共10吨，设乙种水果的进货量为t吨，请你写出这两种水果所获得的销售利润之和W〔万元〕与t〔吨〕之间的函数关系式．并求出这两种水果各进多少吨时获得的销售利润之和最大，最大利润是多少？ 25．〔此题12分〕在△ABC中，ABBC2，ABC120°将△ABC绕点B顺时针，旋转角(0°90°)得△A1BC1，A1B交AC于点E，A1C1分不交AC、BC于D、F两点． 〔1〕如图1，观看并猜想，在旋转过程中，线段EA1与FC有如何样的数量关系？并证明你的结论； D F C C C1 A1 E A D F C1 A1 A E B 〔第25题 图1〕 B 〔第25题 图2〕 〔2〕如图2，当30°时，试判定四边形BC1DA的形状，并讲明理由； 〔3〕在〔2〕的情形下，求ED的长． 26．〔此题14分〕如图，直线l1:y28x与直线l2:y2x16相交于点C，l1、l2分33不交x轴于A、B两点．矩形DEFG的顶点D、E分不在直线l1、l2上，顶点F、G都在x轴上，且点G与点B重合． 〔1〕求△ABC的面积； 〔2〕求矩形DEFG的边DE与EF的长； 〔3〕假设矩形DEFG从原点动身，沿x轴的反方向以每秒1个单位长度的速度平移，设移动时刻为t(0≤t≤12)秒，矩形DEFG与△ABC重叠部分的面积为S，求S关于t的函数关系式，并写出相应的t的取值范畴． yl2 E C D l1A O B F 〔G〕 x 〔第26题〕 2018年山西省初中毕业学业考试试卷 数 学 一、选择题〔每题2分，共20分〕 1021．＞ 2．7.39310 3．答案不唯独，如x1 4．3 5．30 6．210 7．〔9，0〕 8．8 9．3y0 10．3n2 二、选择题〔在以下各小题中，均给出四个备选答案，其中只有一个正确答案，请将正确答案的字母号填入下表相应的空格内，每题3分，共24分〕 题 号 11 12 13 D 14 D 15 B 16 A 17 A 18 B 答 案 D C 三、解答题〔此题共76分〕 19．〔1〕解：原式=x26x9x23x2 ·························································· 〔2分〕 =x6x9x3x2 ····························································· 〔3分〕 =9x7． ·················································································· 〔4分〕 〔2〕解：原式=22xx22 ································································· 〔2分〕 x2x2x2x2 ············································································· 〔3分〕 x2x22 = =1． ··························································································· 〔4分〕 〔3〕解：移项，得x2x3，配方，得x14， ············································· 〔2分〕 ∴x12，∴x11，x23． ···························································· 〔4分〕 〔注：此题还可用公式法，分解因式法求解，请参照给分〕 20．解：〔1〕π2； ··························································································· 〔2分〕 〔2〕答案不唯独，以下提供三种图案． 〔第20题 图2〕 ·································· 〔6分〕 〔注：假如花边图案中四个图案均与差不多图案相同，那么本小题只给2分；未画满四个〝田〞字格的，每缺1个扣1分．〕 21．〔1〕935.7，859.0； ························································································ 〔4分〕 〔2〕解：①2004~2018移动电话年末用户逐年递增． ②2018年末固定电话用户达803.0万户． ··············································· 〔8分〕 2 〔注：答案不唯独，只要符合数据特点即可得分〕 22．解：〔1〕10，50；·························································································· 〔2分〕 〔2〕解：解法一〔树状图〕： 20 10 0 第一次 30 第二次 10 20 30 0 20 30 0 10 30 0 10 20 和 10 20 30 10 30 40 20 30 50 30 40 50 ·················································································································· 〔6分〕 从上图能够看出，共有12种可能结果，其中大于或等于30元共有8种可能结果，因此P〔不低于30元〕=解法二〔列表法〕： 第一次 第二次 0 10 20 30 0 10 20 30 10 10 30 40 20 20 30 50 30 30 40 50 82 ······························································ 〔8分〕 ．123············································································································ 〔6分〕 〔以下过程同〝解法一〞〕 ······································································· 〔8分〕 23．解：分不过A、D作AMBC于M，DGBC于G．过E作D A E 那么四边形AMGD为矩形． EHDG于H，F H AD∥BC,BAD135°，ADC120°．∴B45° ，DCG60°，GDC30°．B G M 〔第23题〕 水深 C ·sinB12在Rt△ABM中，AMAB262． 2∴DG62． ······························································································ 〔3分〕 ·cosEDH2在Rt△DHE中，DHDE33． ······································ 〔6分〕 2∴HGDGDH62-3≈61.411.73≈6.7． ······································· 〔7分〕 答：水深约为6.7米． ···················································································· 〔8分〕 〔其它解法可参照给分〕 24．解：〔1〕由题意，得：ab1.4，a0.1，解得 ········································· 〔2分〕 4a2b2.6．b1.5．2 ∴y乙0.1x1.5x． ········································································· 〔3分〕 〔2〕Wy甲y乙0.310t0.1t21.5t． ∴W0.1t1.2t3． ······································································· 〔5分〕 W0.1t66.6．∴t6时，W有最大值为6.6. ························· 〔7分〕 ∴1064〔吨〕. 答：甲、乙两种水果的进货量分不为4吨和6吨时，获得的销售利润之和最大，最大利润是6.6万元. ········································································· 〔8分〕 25．解：〔1〕EA1FC． ······················································································· 〔1分〕 证明：〔证法一〕 ABBC，AC．22由旋转可知，ABBC1，AC1，ABEC1BF， ∴△ABE≌△C1BF． ··················································· 〔3分〕 ∴BEBF，又BA1BC， ∴BA1BEBCBF．即EA1FC． ······························ 〔4分〕 〔证法二〕 ABBC，AC．由旋转可知，A1C，A1B=CB，而EBCFBA1， ∴△A1BF≌△CBE．···················································· 〔3分〕 ∴BEBF，∴BA1BEBCBF， 即EA1FC． ······························································· 〔4分〕 〔2〕四边形BC1DA是菱形. ······································································ 〔5分〕 证明：A1ABA130°，AC同理AC∥BC1． 11∥AB，∴四边形BC1DA是平行四边形. ················································· 〔7分〕 又ABBC1，∴四边形BC1DA是菱形. ····································· 〔8分〕 C D E G B F 〔3〕〔解法一〕过点E作EGAB于点G，那么AGBG1 ．在Rt△AEG中， A AG12AE3．……〔10分〕 cosAcos30°3A 1C1 由〔2〕知四边形BC1DA是菱形， ∴ADAB2， 2 ················································ 〔12分〕 3．3〔解法二〕ABC120°∴EBC90° ，ABE30°，．2在Rt△EBC中，BEBC ·tanC2tan30°3．32 ·············································· 〔10分〕 EA1BA1BE23．∴EDADAE23AC11∥AB，A1DEA．A1DEA1． ∴EDEA12233． ·······················································〔其它解法可参照给分〕 26．〔1〕解：由23x830，得x4．A点坐标为4，0． 由2x160，得x8．B点坐标为8，0． ∴AB8412． ········································································28由yx，x5，∴C点的坐标为5， ······················33解得6．·········y2x16．y6．∴S11△ABC2AB·yC212636． ·················································· 〔2〕解：∵点D在l281上且xDxB8，yD3838． ∴D点坐标为8，．8 ··········································································又∵点E在l2上且yEyD8，2xE168．xE4． ∴E点坐标为4，．8 ··········································································∴OE844，EF8． ································································ 〔3〕解法一：①当0≤t3时，如图1，矩形DEFG与△ABC重叠部分为五边形CHFGR〔t0时，为四边形CHFG〕．过C作CMAB于M，那么Rt△RGB∽Rt△CMB． y l2 yly2 E l1C D E 2 D l1E lD l1 C R C R R A O F M G B x A F O G M B x F A G O M B x 〔图1〕 〔图2〕 〔图3〕 12分〕 2分〕 3分〕 4分〕 5分〕 6分〕 7分〕 〔 〔 〔 〔 〔 〔 〔 BGRGtRG即∴RG2t． ，，BMCM36 Rt△AFH∽Rt△AMC，112∴SS△ABCS△BRGS△AFH36t2t8t8t． 223421644即St ·························································· 〔10分〕 t．333∴