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  • 炎德英才大联考2024年普通高等学校招生考试考前演练四英语试题

    形式。

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    10.某人在11次本卷考查:1随机变量及其分布:2.满分150分考试时间120分钟若P(X=kA.7合分人复分人第Ⅱ卷总分赛的学生中随机选择一名,他的竞赛成绩为A等的概率为()11.在某独立重事件A发生题号第I卷(附:若X~N(4,o2),则P(4-G≤X≤4+o)=0.6827,p∈(0,1).者得分第I卷选择题(共60分)P(u-20≤X≤u+2o)=0.9545每小题5分,共60分.在每小题给P(μ-30≤X≤u+3o)=0.9973)发生的次数是()本大题共12小题,A.0.0455B.0.0214C.0.0428一、出的四个选项中,只有一项是符合题目要求的。选择题:D.0.02275已知随机变量X的分布列为PX=)=。(1=1,2,3,4,则6.先后抛掷一颗质地均匀的骰子两次,观察向上的点数.在第一次A.pE(XB.(1-pa向上的点数为奇数的条件下,两次点数和不大于7的概率为1.()C.E(ZP2≤X<4)=(.、7D.9A13718B.12.3D.「D(B.51001012.甲、乙、A.22.在四次独立重复试验中,事件A在每次试验中发生的概率相同,7.某次国际象棋比赛规定,胜一局得3分,一局得1分,负一局次传球时得0分,某参赛队员比赛一局胜的概率为a,局的概率为b,负人,则若事件4至少发生一次的概率为65则事件A发生次数5的期81的概率为c(a,b,c∈[0,1)),已知他比赛一局得分的数学期望为A.1-1,则ab的最大值为(望是(B.c.8D.832D.1463.一台机器生产某种产品,如果生产出一件甲等品可获利50元,生8.某群体中的每位成员使用移动支付的概率都为P,各成员的支付产出一件乙等品可获利30元,生产一件次品,要赔20元,已知方式相互独立,设X为该群体的10位成员中使用移动支付的人阳这台机器生产出甲等、乙等和次品的概率分别为0.6、0.3和0.1,数,D(X)=2.4,且P(X=4)

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    形结合思想,函数与方程思想等;考查直观想象,逻辑推理,数学运算等核心素养;体现基础性和综合性.满分12分.解法一:(1)如图1,取AB中点O,连接PO,CO因为PA=PB=V2,AB=2,所以PO⊥AB,PO=1,BO=1.又因为ABCD是菱形,∠ABC=60°,所以CO⊥AB,CO=V3因为PC=2,所以PC2=P02+C0,所以P0⊥C0.又因为ABC面ABCD,COC面ABCD,AB∩CO=O,所以P0L面ABCD..2分因为AD∥BC,BCC面PBC,AD文面PBC,所以AD∥面PBC,所以VD-Pac=VAPc=V,-ABC=x1x5x4=5433分因为VM-PBc==VD-P8’.4分62所以点M到面PBC的距离是点D到面PBC的距离的2,所以PM=MD.5分(图1)(图2)(2)由(1)知,B0⊥C0,PO⊥B0,P0⊥C0,如图2,以0为坐标原点,OC,OB,OP的方向分别为x轴,y轴,z轴正方向建立空间直角坐标系,6分则A(0,-1,0),B(0,10),C(V3,0.0,D(3,-2,0,P(0,01),所以M则c-,c-版-m-点-a.西-9-引因为Q∈AP,设AQ=AP=(0,,),则C0=A0-AC=-V3,2-1,),数学参考答案及评分细则第8页(共20页)

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    15.直线I:y=2x和2:y=kx+1与x轴围成的三角形是等腰三角形,写出满足条件的k的两个可能取值:和.(写对一个得3分,写对两个得5分)16.在同-面直角坐标系中,P,Q分别是函数f(x)=axe'-n(ax)和g(x)=2n(x-1图象上的动点,若对任意a>0,有1PQ1≥m恒成立,则实数m的最大值为四、解答题:本题共6小题,共70分。解答应写出文字说明、证明过程或演算步骤。17.(10分)》记数列{an}的前n项和为Sn,对任意neN',有Sn=n(an+n-1).(1)证明:{an}是等差数列;(2)若当且仅当n=7时,S,取得最大值,求a1的取值范围.18.(12分)设△ABC的内角A,B,C所对的边分别为a,b,c,且有2sin(B+)=6+C6a(1)求角A;(2)若BC边上的高h=4a,求cosBeosC.19.(12分)》如图,在边长为4的正三角形ABC中,E,F分别为边AB,AC的中点.将△AEF沿EF翻折至△AEF,得到四棱锥A,-EFCB,P为A,C的中点,(1)证明:FP∥面A,BE:(2)若面A,EF⊥面EFCB,求直线A,F与面BFP所成的角的正弦值数学试卷第4页(共5页)

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    C.|x-2|<|2-y二、填空题:本题共4小题,每小题5分,共20分。D.|x-3<|3-yA尖直面早特(3.已知图数n=T2若2三6.23.2=2,g2=1.则1(2)店D14.请写出一个与y轴利直线)一:都相切的圆的方遥B15.如图,正方体ARCD AB,CD的校长为4,则三棱锥CDB1一A1BC和三棱锥A,-AB,D,公共部分的体积为16.在一个正三角形的三边上,分别取个距顶点最近的十等分点,连接形成的三角形也为正角形如图1所示图中共有2个正三角形),然后在拉小的正三角形中,以同样的方式形成一个更小的正三角形,如此重复多次,可得到如图2所示的优美图形(图中共有1个正三角形),这个过程称为迭代在边长为243的正三角形的三边上,分别取一个三等分点,连接成一个较小的正三角形,然后迭代得到如图3所示的图形(图中共有10个正三角形),其中最小的正三角形的面积为外立加林图1图2图3三、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答.第22、23题为选考题,考生根据要求作答(一)必考题:共60分17.(12分)已知△ABC的内角A,B,C的对边分别为a,bc,且cos2B-cos2C-sinA·(sinAsin B).(1)求角C的大小;(2)若CDLAB于点D,CD=3,求△ABC的面积的最小值.18.(12分)如图,矩形ABCD所在的面垂直于半圆所在的面,AB=2AD=2a,点E是半圆O上异于A,B的点(1)证明:EA⊥EC.全国100所名校最新高考冲刺卷第3页(共5页)【23·高考样卷·数学文科(二)一GQN·A】

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    ∠FBD=∠ABD+∠ABF:∠BFD=∠DCB+∠CBF∠ABF=∠CBF,.∠FBD=∠BFD.BD=FD…10分六、(本题满分12分)2L.解:(1)总人数为9÷18%=50(人)50-9-14-6=21(人).B级人数为21人,统计图略…4分(2)100.8°78.5…8分(3)650x(18%+42%)=390(人)》“.九年级达到良好及良好以上的学生人数约为390人…12分七、(本题满分12分)】22.解:(1y=x2+24-3=(x+1P-4顶点为(-1,-4)y=x24-3=-(x+22+1顶点为(-2,1)设根活数为y则广法村相你。,∴.根函数为y=-5x-9…4分(2①y=x24mx+4m34m+1-(x-2mP4m+1顶点为(2m,-4m+1)-4m+1=-2x2m+1∴.x=2m时,y=-2x+1:,满足二次函数C的所有二次函数的根函数为=-2x+1…8分②x=-3时,Jy=(-324mx(-3)+4m2-4m+1-4m248m+10,点P坐标为(-3,4m2+8m+10)设-4m+8m+10-4(m+1)+6.当m=-1时,点P纵坐标最小值为6即点P到x轴的最小距离为6,此时m=-1.…12分八、(本题满分14分)23.(1)证明:四边形ABCD是正方形,∠DBE=45△EBG是等腰直角三角形,能-妥:CD∥RCP/DE人四边形DEC为行西边形.EF-DC2:△BCD是等膜直角三角形÷品--孚∴配品BDBD2又:∠GEF=LEBD=45.∴.△BDE△EFG…4分(2)解:由(1)知△BDE∽△EFG∴,∠EFG=∠BDEg四边形DEFC为行四边形.∠EFC=∠EDC÷∠CFG=∠BDC=45°8分(3懈GE-V2EF-CDCD-②BE2BD 2…品∠FEG=∠DBE=135.△GEFM△EBD器品:H为BC中点∴,△DCH≌△EBH.BE=CD-4:..GE-2V2 DE=VADTAE=V4+8=4V5器-部2vV2 GrGF-2V1044V5DG-DB+BG-4V7+2V2-6V2 PC-2V10V5…14分DG 6V2 3-2-

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    参考答案1.B解析:易知C={1,2},其子集个数为22=4,故选B.2.A解析:根据题意知,表示向量A的复数为1十5i,所以在复面上所对应的点为(1,5),位于第一象限.故选A.3.D解析:根据雷达图,甲同学按照科目综合指标值从高到低顺序为物理、历史(化学)、地理、生物、政治,乙同学按照科目综合指标值从高到低顺序为历史、物理(政治)、地理、生物、化学.根据新高考选科模式规则,选考科目最合理的是:甲应选物理、化学、地理;乙应选历史、政治、地理.故选D.4.B解析:角a的始边与x轴非负半轴重合,终边过点A(a,2a)(其中a≠0),tan a-2,cos 2a+cos'a-2cos'asin'a=2-tan'a2sina+cos2 x tana+15·故选B.A解析0a=3一2.u:2了5=-3-2,4=二2y5-2√3+1-3-2√3+1所以{am}是周期为3的数列,因为2023=3×674+1,所以a2o23=a1=1.故选A.6.C解析:由A和B是圆C:(x一1)2十y2=1上的两点,且AB=√2,则∠ACB=2不妨设A(1十cos6,sin6,则B(1+cos(0+5),sin(6+5),即B(1-sin0,cos6),PA=(cos 0-3,sin 0-1),PB=(-3-sin 0,cos 0-1),3 PA-PB=(sin 0++3cos 0-6,3sin 0-cos 0-2),|3PA-Pi12=(sin0+3cos0-6)2+(3sin0-cos0-2)2=50-8(3sin0+4cos0)=50-40sin(0+o),又sin(0+o)∈[-1,1],即当sin(0+o)=-1时,3PA-PB12的最大值为90,则|3P才-PB的最大值为3√10,故选C.7.C解析:可以直接求导证明如下结构的不等式:e>1十x(x≠0),x>sinx(x>0),xeo.1>sin(eo.1),从而a>c,对于b,c:l>eo.1>l-0.1=0.9,sin(e-o.1)>sin0.9,从而c>b,所以a>c>b,故选C.22023届高三“一起考”大联考(压轴卷)·数学参考答案

  • 炎德英才大联考2024年普通高等学校招生考试考前演练四英语试题

    北京市朝阳区2022~2023学年度第一学期期末质量检测高三英语2023.1(考试时间90分钟满分100分)》本试卷共10页。考生务必在答题卡指定区域作答,在试卷上作答无效。第一部分知识运用(共两节,30分)第一节(共10小题;每小题1.5分,共15分)阅读下面短文,掌握其大意,从每题所给的A、B、C、D四个选项中,选出最佳选项,并在答题卡上将该项涂黑。Early in my teaching career,I heard countless make-believe stories for unfinished homework.Then I grew less trusting and quitted 1 any excuse.When the students at my new schooldidn't finish homework,I never asked why.Instead,I just sighed loudly and recorded a zero inthe grade book.I soon gained the 2 I thought I wanted.One day,Anthony approached me..“Could I talk to you?”he asked_3_·“I know yousaid no excuse,but I don't want you to think I'm 4 because I often come without myhomework."He then looked up at me for the first time."It's just that...my dad moved out,andmy mom works at night,so I have to look after my little brothers.Often,they cry a lot,and it'shard for me to_5_”I was about to ask why he didn't tell me earlier when I suddenly realized why.So I changedthe 6."Would it help if you stayed after school and worked on it before you go home?"He nodded hard.Anthony became the first student in my after-school study session.Several days later,Terrelljoined him,followed by Sandy and Randy.Before long,I had a room full of students.Theirstories were not amusing,but all very 7:The power company 8 our lights because we couldn't pay the bill.My dad says schoolwork is just a waste of time.·We don't have any paper i讥the house.I thus discovered not all kids come from families that are 9.Not all kids have a quietbedroom with a desk and study light.Some don't even have home.Most importantly,I learnedthat“I'll listen'”10 better than“No excuse!'”1.A.inventingB.findingC.acceptingD.offering2.A.reputationB.benefitC.experienceD.praise3.A.loudlyB.shylyC.curiouslyD.eagerly4.A.p00rB.meanC.lazyD.weak5.A.behaveB.concentrateC.struggleD.compete高三英语试卷第1页(共10页)

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    【第40期】第3章3.1~3.3同步测试题-、1.D2.C3.B4.D5.A6.C7.B8.B9.C10.B二、11.(5,2)12.-1013.(-3,4)14.315.(1.0)16.217.(3.3)或(-3.3)18.(-3,-2三、19.因为点P在过.点A(2,-3)且与x轴行的直线上,所以点P的纵坐标与点A的纵坐标均为-3,即m-1=-3.解得m=-2.故点P的坐标为(0,-3).20.(1)点M(2.3),N(-3,2),L(0,-2),0(0,0),P(2,-2).(2)点A,B,C,D的位置如图所示A-10ED第20题图21.(1)因为点A在y轴上,所以横坐标为0,即3a-5=0.解得a=号将a=号代入.可得点A的坐标为(0,》(2)依题意,有3a-5=a+1,即3a-5=a+1或3a-5=-(a+1).①当3a-5=a+1时,得a=3,此时点4的坐标为(4,4);②当3a-5=-(a+1)时,得a=1.此时点4的坐标为(-2,2).因为点A在y轴的左侧,所以点A的横坐标为负数.所以a=1,点A的坐标为(-2,2).22.(1)△A'B'C如图所示.点B(-4,1),点C(-1,-1).(2)点p'的坐标为(a-5,b-2).y米AB2B4-3102第22题图23.台风的起始位置为(4,-8),B市的位置为(4.0)当B市受影响时,台风的位置是点C(4,-3).所以100×(8-3)÷10=50(时).所以经过50小时后,B市将受到台风的影响,画图略.

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    高三数学考试参考答案1.A【解析】本题考查集合的运算,考查数学运算的核心素养,因为A={x00,所以t∈(3,4).6.D【解析】本题考查解三角形的知识,考查数学运算的核心素养,因为bcos A=a(W3-cosB),所以sin Bcos A=√3sinA-sin Acos B,移项得sin Bcos A十sin Acos B=√3sinA,即sinC=√3sinA,所以c=√3a=2√3,7.C【解析】本题考查抽象函数的求值,考查数学抽象的核心素养!因为f(x+y)=f(x)+f(y),所以f(0+0)=f(0)+f(0)=2f(0),即f(0)=0.所以fln2023)+fln2023)=f1n2023-ln2023)=f0)=0.8.B【解析】本题考查数学文化与等比数列的求和,考查数学抽象与数学运算的核心素养由题意,若正整数m≤6,且与6“不互质,则这个数为偶数或3的倍数,共有号×6个,所以(6”)=号×6”=2X6"1,即数列{(6”)}是首项为2,公比为6的等比数列,所以S2=32g-号0t-10.9.AC【解析】本题考查直线与圆的位置关系,考查数学运算的核心素养对于A,因为圆心C(1,2)在直线y=kx十1上,所以2=k+1,解得k=1,A正确;【高三数学·参考答案第1页(共8页)】531C

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    13:10|4.9K/s2⊙o56Ho49ll☑0〈数学试卷.pdf●●●23.(本题13分)如图,已知二次函数y=-x+bx+c的图象交x轴寸点A(-1,0),B(5,0),交y轴于点C(1)求这个二次函数的表达式:(2)如图1,点M从点B出发,以每秒V2个单位长度的速度沿线段BC向点C运动,点N从点O出发,以每秒1个单位长度的速度沿线段OB向点B运动,点M,N同时出发.设运动时间为t秒(0<1<5).当t为何值时,△BMN的面积最大?最大面积是多少?(3)已知P是抛物线上一点,在直线BC上是否存在点Q,使以A,C,P,Q为顶点的四边形是行四边形?若存在,直接写出点Q坐标:若不存在,请说明理由.备用图718泪OD目录模式播放三▣

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    22.长方体的一个顶点上的三条棱长分别是3,4,5且它的8个顶点都在同一个球面上,则这个球的表面积为5>.三、解答题:本大题共3小题,共30分.解答应写出文字说明、证明过程或演算步骤24.(23.(本小题满分10分)某网校推出试听的收费标准为每课时100元,现推出学员优惠活动,具体收费标准如下(每次听课1课时):第n次课第1次课第2次课第3次课第4次课或之后收费比例0.90.80.70.6现随机抽取100位学员并统计他们的听课次数,得到数据如下:听课课时数1课时2课时3课时不少于4课时频数50201020假设该网校的成本为每课时50元。(1)估计1位学员听课三次及以上的概率:(2)求一位学员听课4课时,该网校所获得的均利润的家中如0对,n日,前,盖阴动数激1。顺为第京+心1入是,数学·模拟卷四第4页(共5页)

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    参考答案学生月人书所以g(x)mx=g(3),即7m一6<0,fa2-4(3-a)≥0,1a≥2或a≤-6,所以m<号,所以00,脚x=-受≥2,g(2)≥0,又因为m(x2-x十1)-6<0在x∈[1,3]上恒成立,fa2-4(3-a)≥0,a≥2或a≤-6,所以m<2-x+在x[1,3]上恒成立.6即-号≥2,可得a≤-4,7+a≥0,a≥-7.6令J-2-x+1∴.-7≤a≤-6,综上,实数a的取值范围是[一7,2].因为函数y2-x+166在[1,3]上的最小值为9,(3)令h(a)=xa+x2+3.(-)+当a∈[4,6]时,h(a)≥0恒成立.所以只需m<号即可。只音8甲计十:所以m的取值花国是(-0,号)】解得x≤-3-√6或x≥-3十√6.∴.实数x的取值范围是(一∞,-3一√6]U[一3十6,十∞).例8[解析]由f(x)=x2+(m-4)x十4-2m第2章函数=(x一2)m十x2-4x+4,第6讲函数及其表示令g(m)=(x-2)m十x2-4x十4.【基础检测】由题意知在[一1,1]上,g(m)的值恒大于零g1)=(-2)X(-1)+2-4x+40,1.(1)×(2)×(3)/·(4)×2.[-3,6)1g(1)=(x-2)+x2-4x十4>0,3.[-3,0]U[2,3][1,5][1,2)U(4,5]解得x<1或x>3.4.D[解析]因为函数f(x)的定义域为[0,2],故当x∈(-∞,1)U(3,十o∞)时,对任意的m∈[-1,1],函数f(x)的所以0≤2x≤2,解得0≤x≤1,值恒大于零,所以函数f(2x)的定义域为{x|0≤x≤1.例9[解析]f(x)=x2-2x十3=(x-1)2+2,5.x2-1(x≥0)[解析]令1=√:,则≥0,x=2,所以f(t)=2-1(t当x∈[1,4]时,[f(x)]min=f1)=2,≥0),所以f(x)=x2-1(x≥0).[g(x)]m=g(4)=2十m.6.5士1[解析]f1)=1X(1+4)=5.当a≥0时,由f(a)=a2+4a由题意得,[f(x)]mn>[g(x)门mx,=5得a=1;当a<0时,由f(a)=a2-4a=5得a=-1.则2>2+m.综上可知a=土1..m<0.【知识要点】例10[解析]当0≤x≤3时,[f(x)]m血=f(0)=0,1.非空数集任意唯一确定∫:A→B当1<<2时,[g(x]m-g(2)-}-m,2.(1)定义域函数值值域(2)定义域对应关系值域电题意得fxl≥[gx门m0≥号-m(3)解析法图象法列表法3.对应关系并集并集m>≥【关键能力】训练巩固例(1)B[解析]由集合M={x1-1≤x≤3),N={y10≤y≤3),7.[解析](1),当x∈R时,x2+ax十3-a≥0恒成立,可排除A,D.由函数的定义知,每一个x只能对应唯一一个y,而C选项需△=a2-4(3-a)≤0,即a2+4a-12≤0,中当一1≤x<3时,每一个x都有两个y与之对应,不符合函数的定义,.实数a的取值范围是[-6,2].故排除C.(2)当x∈[-2,2]时,设g(x)=x2+a.x十3-a≥0,分如下三种情况讨(2)BC[解析]A.f(x)=x一1的定义域为R,论(如图所示):g(x)=-1的定义域为(-o∞,0)U(0,十∞),s(a)故f(x)≠g(x),A错误;/g(x)B.f(x)=log221=x,g(x)=故f(x)=g(x),B正确;20202C.f(x)=|x|,g(x)=x2,图1图2:|x|=√,且f(x)与g(x)定义域相同,∴f(x)=g(x),C正确;g()D.f(x)=nx2的定义域为(-o,0)U(0,十∞),g(x)=2nx的定义域为(0,十∞),故f(x)≠g(x),D错误。-202训练巩固1.C[解析]①函数y=2x十3的定义城为R,值域也为R,即定义城和图3①如图1,当g(x)的图象恒在x轴上方且满足条件时,值城相同;②函数y=上的定义城为(-∞,0)U(0,十c0),值城也为有△=a2-4(3-a)≤0,即-6≤a≤2.(一∞,0)U(0,十o),即定义域和值域相同;③指数函数y=2的定义②如图2,g(x)的图象与x轴有交点,城为R,值城为(0,十∞),即定义城和值域不同;④幂函数y=x意的定义但当x∈[-2,+co)时,g(x)≥0,域为[0,十∞),值域也为[0,十c∞),即定义域和值域相同.故选C.4>0,x+1,-1≤x<0即x=-≤-2,2.f(x)=l-zz,0Kx≤2g(-2)≥0,[解析]由题图可知,当一1≤x<0时,f(x)=x十1;当0≤r≤2时,f(x)=-x507

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    第9章统计学生用书心训练巩固临界值表:6.某机构为了了解不同年龄的人对一款智能a0.050.010.001家电的评价,随机选取了50名购买该家电Ta3.8416.63510.828的消费者,让他们根据实际使用体验进行评分.(1)设消费者的年龄为x,对该款智能家电的评分为y.若根据统计数据,用最小二乘法得到y关于x的经验回归方程为y=1.2x十40,且年龄x的方差为s2=14.4,评分y的方差为s号=22.5.求y与x的相关系数r,并据此判断对该款智能家电的评分与年龄的相关性强弱.(2)按照一定的标准,将50名消费者的年龄划分为“青年”和“中老年”,评分划分为“好评”和“差评”,整理得到如下数据,请依据α=0.01的独立性检验,能否认为对该智能家电的评价与年龄有关联.评价年龄好评差评青年816中老年206附:经验回归直线y=x十a的斜率b=0相关系数r=2(x,-x0(0-),独立性检验中2x-2-n(ad-bc)2的t=a+a十c)6+d0c+D其中n=a+6+c+d.303

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    参考答滨学生月书x(0,2)2(2,+o∞)于S点,正三角形的边长为x>0,则0G=子×县x=侣f(x)+0f(x)元1n2-1¥故f(x)在定义域上的极大值为f(x)机大值=f(2)=1n2-1,无极小值.(2)由(1)知,函数f(x)的定义域为(0,十∞),当a≤0时,f(x)>0在(0,十∞)上恒成立,则函数在(0,十∞)上单调递增,此时函数在定义域上无极值点;当>0时,若x(0,),则fe>0,若z(日,+),则f田)<0,s0==v50-远-√-得)-()√5(5-鼓通数在x=日处有板大值。三校维的体积V=号S0版·h=子×92×√5(5-号)=综上可知,当a≤0时,函数f(x)无极值点,当>0时,西数)=:)有-个报大值点,且为x=合8[解析](:f✉)=士+红-a(>0,设n)=5r-得,>0,则e=202-54,又f(x)在(0,十∞)上单调递增,令n(a)=0,即4x-号=0,得x=43,易知n(x)在x=45处取得最所以包有fx20,甲士十红-。≥0(>0)框底立,a<(+)√3大值.面叶上≥2√工=2,当且收当=1时取等号,所以a<2Vu=爱×48X5=4压,即函数f(x)在(0,十o∞)上为单调增函数时,a的取值范围是(一o∞,2]】3[解折]af)=+ar一a+D=a-D-卫(>0,2f)-fa)=lh妥+7(d-)-a-,①当a≤0时,ax-1<0,故当00,f(x)单调递增,当x>1时,f(x)<0,f(x)单调递减,此时f(x)在x=1处有极大值;又“fe)=}+红-a=-2出(a>0.x②当0a<1时,合>1,令f)=0,解得x=日=1所以x1,x2是方程x2一ax十1=0的两个实根,故当00,f(x)单调递增,由韦达定理得:x1十2=a,x1x2=1,当1K<日时,fx)<0,f)单将造藏,“fx)-fx)=n+是(-x)-a(-)当>是时,f:)>0,x)单调递增此时f(x)在x=1处有极大值;=n器-(贤-》③当a=1时,f(x)≥0,f(x)在定义域内单调递增,无极大值;④当>1时,甲0<日<1,令了x)=0,解得x=日=1.设=要,故当0<<日时,f:)>0,)单调逢增。令ae)=h-(:-)②0.当1时,∫(x)>0,f(x)单调递增,此时f(x)在x=1处有极小值综上所述,当a<1时,f(x)在x=1处有极大值,h(t)在[E,十∞)上是减函数,即a的取值范围是(-∞,1).a0<)=2(1-E+9),(2)由(可知,当0Ka<1时,M=fID.N=日),故)-a)曲最大值为号(1-+)】当1时,M=(日)N=f.B组题所以1M-N=(日)-f)1.C[解折]由题意可知,x1<00,所以-2+2=品十,则理-2西-十2-2,-(&-a-)-(-号-)令ux)=+2x-2>0且x≠1,ux>0,=血日-云+号@>0且a≠1011.①当00,②当x>1时,可知u(x)>0,4(x)=2nx2+lnx-1-(In z)2则g)=-士+2+=2+1-业0,2x22x22血-)也x+,令(x)=0,则=,=。(舍去),若11单调递增,<√e,则u'(x)<0,若x>√E,则'(x)>0,则x=√E时取得极小值u(W)于是|M-NI=lg(a)l,=4Ve-2,也为最小值,所以u(x)≥u(WE),即x2一2x1≥4Ve-2,所以所以M-N训=lg@)1在a=合或a=2处取得最大值,g(合)x2-21的最小值为4ve-2.故选C.是-h2,g21=是-ln2,32.4√5cm3[解析]如下图,连接DO交BC于点G,设D,E,F重合由于a>0且a≠1,lM-N|=g(a)川>g(1)=0,639

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    基础题与中考新考法·八年级·上·数学第3课时用“ASA”或“AAS”判定三角形全等(共424考:单独87考,在几何综合题、函数综合题等中涉及337考)】教材知识知识点①三角形全等的基本事实:角边角69考·重点点拔1.基本事实:两角和它们的夹边分别相等的两个三角形全等,注意:(1)“ASA”与“AAS”简记为ASA(或角边角)的区别:“ASA”中边是两角2.几何语言:如图,在△ABC和△DEF中,的夹边,“AAS”中边是其中∠B=∠E,一角的对边;(2)“ASA”与“AAS”的联BC=EF,系:由三角形内角和定理可∠C=∠F,知,“AAS”可由“ASA”推导..△ABC≌△DEF(ASA).得出.知识点(2三角形全等的基本事实:角角边5考1.基本事实:两角和其中一个角的对边分别相等的两个三角形全等.简记为AAS(或角角边).》注意2.几何语言:如图,在△ABC和△DEF中,∠B=∠E,∠C=∠F,AC=DF,..△ABC≌△DEF(AAS).重难题解法》答案见本册P68●“截长补短法”构造全等三角形8考·现学活用·练-1如图,在四边形AB例一题多?解法如图,在△ABC中,∠A=100°,∠ABC=40°,CD中,∠ABC=∠ADC=BD分∠ABC,延长BD至点E,连接CE,且∠DCE=∠DCB,90°,AD=DC,若BD=8cm,求证:BC=AB+CE.则四边形ABCD的面积为规范解题证法一:截长法练-1题图例题图18

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

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  • 炎德英才大联考2024年普通高等学校招生考试考前演练四英语试题

    10:47|12.1K/s2⊙HD 5G HD 4G1615b2f35bff0ac5a4711a28d16222f06..文件预览1/6最新中考模拟训练试题·英语参考答案山西省2023年最新中考模拟训练试题·英语参考答案(四~六)》(四)一、情景反应本题共5个小题,每小题你将听到一组对话。请你从每小题所给的A、B、C三幅图片中,选出与你所听到的信息相关联的一项,并在答题卡上将该项涂黑。1.W:Do you like playing football?M:Yes,I do.2.W:Who's the woman with a pair of glasses?M:Oh,she is my English teacher.3.W:What will you do on the weekend?M:I will play the drums with my brother.4.W:Don't throw your bottle here.M:Sorry.5.W:You look tired,Tom.M:We ran for half an hour on the playground just now.二、对话理解本题共5个小题,每小题你将听到一组对话和一个问题。请你从每小题所给的A、B、C三个选项中,选出一个最佳选项,并在答题卡上将该项涂黑。6.M:Can you play the piano,Alice?W:No,but my sister can.Q:Who can play the piano?7.W:It has rained for three days,but it will be sunny tomorrow.M:I'm so glad to hear that because I'd like to have a picnic tomorrow.Q:What will the weather probably be like tomorrow?8.W:I am happy that we will go to Hainan together next week.M:So am I.Oh,we should buy something in the mall today.Q:Where will they go today?9.W:What's the matter,David?M:I have a stomachache.Q:What's the matter with David?10.W:I heard that Victor failed the driver's test last week.Is it true?M:Yes,it is.And it's really a pity.He had practiced it for a long time.But he feltnervous when he was having the test.Q:Why did Victor fail the driver's test?三、语篇理解本题你将听到一篇短文。请你根据短文内容和所提出的5个问题,从每小题所给的A、B、C三个选项中,选出一个最佳选项,并在答题卡上将该项涂黑。It is reported that a girl made a short video which swept the screen.In the video,youcould see a young Chinese girl playing guzheng on the street of Paris,the capital city ofFrance.She wore beautiful traditional Chinese clothes,performing Chinese traditional songs.Not only the local people but also the visitors from all over the world couldn't help stoping andlistening to her beautiful guzheng performance.She has become a special scenery in Paris.The girl's name is Peng Jingxuan.In 2010,she went to college in Wuhan for professionaltraining.In 2017,she went to Paris for further education.One day,as Peng watched a bandperforming in Paris,she had mixed feelings.She came up with the idea of spreading tradition-al Chinese music to the world.In 2018,she managed to take her guzheng to Paris and startedto play guzheng on the street.She also patiently taught foreigners to pronounce Chinese char-(7)【SHX】最新中考模拟训练试题·英语参考答案acters.The people were moved,just as an American visitor said in the video,"Although Ican't speak Chinese,I am moved by the wonderful Chinese music."The Chinese girl won the praise of the world for her spreading our national culture.四、听力填空本题你将听到一篇短文。请你根据所听内容,完成下面的句子,并将获取的信息填到答题卡相应的位置上。每空一词Do you know the "green"car?In 1891,William Morrison built the first working electriccar in the US.The "green"cars have become popular in recent years.They are called electriccars.In fact,they have a history of nearly more than 100 years.To begin with,although some electric cars could just run 40 miles at a time,people coulddrive them instead of walking.In 1908,the gas-powered car appeared.People stopped buyingelectric cars because gas cars could travel longer.But after over 60 years,the world was shortof the gas.So another electric car came out again.Although it was a pity that it had safetyproblems,it really helped people solve the problem of being short of gas.Nowadays,people all over the world are developing electric cars and want to use themmore in our life.China also makes great efforts.In many cities,the taxis were all changedinto electric cars,just because it can prevent the air pollution and it is helpful for三

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    三、填空题:本题共4小题,每小题5分,共20分」ars d=-13.已知等差数列{.}的前n项和为S.,且a,=-1,2S,=7a,若将4,a,a,a6/0去掉一项后,利下三项依次为等比数列包}的前三项,则6为又。2a62沙新接海因14.程为)015第三届中非经贸博览会于2023年6月29日在湖南长沙举行,组委会准备安排甲2--700.51.522.53.5x乙等5名工作人员去A,B,C,D这4所场馆担任服务工作,每个场馆至少安(1)若从水库中随机捕捞一条鱼,求鱼的重量在[2.5,3.5]内的概率;排1人,其中甲、乙不能安排在同一场馆,且乙不能安排到A场馆,则不同的安C2(2)从捕捞的100条中随机挑出6条鱼测量体重,6条鱼的重量情况如表:排方法种数为82叱22重量范围(单位:kg)[0.5,1.5)1.5,2.5)[2.5,3.516.正三棱柱ABC-AB,C的底面边长是4,侧棱长是6,点M,N分别为CC,AB的条数13中点,若点P是侧面BCCB,上一点,且PN∥面AB,M,则线段PN的最小值为22①为了进一步了解鱼的生理指标情况,从6条鱼中随机选出3条,记随机边s1A8)污B)出的3条鱼中体重在[2.5,3.5]内的条数为X,求随机变量X的分布列和数号期望;四、解答题:本题共6小题,共0分解答应出文字说界理过政演算步果八了化05(十官资说缝石标言2条为了调整生态结构.能进种群的优化,②若将选剩下的94条鱼称重微标记后立即放生,两周后又随机捕捞1000务7.(10分况△48C的内角4.B,C约对边分别。餐动-2e…C-君)70(OC十0客福研在2:写司丙的鱼的数的4进行出告,底待算水库中鱼的2b((1)求A;losccostesincsinsV3n225a(5数以及应捕捞体重在[2.5,3.5]内的鱼的条数(2)设AB的中点为D,若CD=a,且b-c=1,求△ABC的周长3sinB=2sinAlosine.L18.(12分)正态分布有极其广泛的实际背景,生产与科学实验中很多随机变量的概率19.(12分如图,在四楼维P-ABCD中,底面4BCD为菱形,∠ABC-,PA1底面ABCDAB=PA,E是PC上任一点,AC∩BD=O分布都可以近似地用正态分布来描述.例如,同一种生物体的身长、体重等指标.随(H)求证:面EBD⊥面PAC;着人们对环境保护意识不断增强和环保工作的积极推进,环境问题得到有效解决.为了调查某水库的环境保护情况,在水库中随机捕捞了100条鱼称重.经整(2)四棱锥P-ABCD的体积为Y,三棱锥B-CED的理分析后发现,鱼的重量x(单位:kg)近似服从正布x~N么,。),如图所体积为,者合-子求直线P9与面6D所示,已知P(x<0.5)=0.05,P(x≤1.5)=0.29成角的正弦值卷第4页L05A)广+(5升-COSA-数学试题卷第5页共6页3C05闪千Uo2

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    LHtilPF=IPC=7--11分2即当PF=2时,二面角F-AB-C的余弦值为25网-12分1720.解:(1)fw=2e2-2a+10e+2a=2(e-le-a)11分①a=1时,由f'(x)20,f(x)在(-0,oo)上单调递增.-2分②a>1时,由f'(x)>0得x<0或x>na,f'(x)<0得00得x0,f'(x)<0得lna1时,由(1)得f(9在(0,l血a上单调递减,在(血a,+∞)上单调递增所以对任意x>0,f(x)zf(Ina)=e2a-2(a+l)eh@+2aln a=a2-2a(a+1)+2alna=2alna-a2-2a------8分Ag(a)=2alna-a2-2a (a>1),则g'(a)=21+lna))-2a-2=2(na-a)<0.10分所以g(a)在(1,+oo)上单调递减,g(a)1,关于x的方程f(x)=k恒有正数解,所以k≥-3.-12分21解:(D因为0=1+m=克所以m=分1分)9stn2ax-1-co2o+,所以0)=sin(2wx+月.3分2又因为函数f(x)的最小正周期为π,所以w=1,所以:f)=sin(2x+君.5分当2x+君=2纸-受,keZ,即x=a-号,kez时,0)am=-1所以函数f(x)的最小值为-1,.-6分2》令m2x+合=0,则2x+名-,keZ,所以x=经-合keZ.-8分66试卷集68共8页

  • 炎德英才大联考2024年普通高等学校招生考试考前演练四英语试题

    C.A kind heart.D.A charitable deed.6.What can be inferred from the "cupcake"experiment?A.Participants involved preferred sweeter cupcakes.B.Random kindness has much greater effects than expected.C.Extra ingredients make the tasty treat more special and unique.D.Kindness performers didn't realize it could make people feel good.7.What is the main idea of the text?A.Small kindness matters.B.Small kindness is a choice.C.Small acts are life-changing.D.Small decisions have a big impact.Just 130 years ago scientists weren't sure if any planets existed outside our solar system.Now they've detected more than 5,000 of them.But as astronomers have calculated theseexoplanets'sizes,a strange gap has emerged.There are plenty of "super-Earths"out there-rocky globes about 1.4 times wider than Earth.And there are lots of"mini-Neptunes"roughly2.4 times Earth's width.But very few planets fall in between;it's almost like most worldswere sized using one of two cookie cutters.A new model published in The Astrophysical Jour-nal Letters offers a fresh answer to why this is so:it's all about collisions ()Previous theories about the planetary“radius(半径)gap”suggested that extreme heatminify certain planets,says Rice University physicist Andre Iidoro,lead author of the new1study Planets tend to move coetotheir host stars over time,he says.This makes relativelylight planets sim down faster as rising heat takes away their outer gases,the thinking goes,whereas heavier planets have enough gravitytohold these gases and maintain theirseidrork challengesthis heat-basedxtion,suggesting the gap resmpnery collisions instead.AHis team ran computer simulations (based theories of how planetary systm全国100所名校最新高考冲刺卷第4页(共12页)【23·(新高考)高考样卷·英语(一)一SD】全国100所白X四

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    21.(本小题满分12分)如图,在四棱锥P-ABCD中,△ABD是边长为2的正三角形,BC⊥CD,BC=CD,PD⊥AB,面PBD⊥面ABCD(I)求证:PD⊥面ABCD:(2)若PD=1,求二面角C-PB-D的面角的余弦值.22.(本小题满分12分)已知函数f(x)=2-2,g(x)=log2x.()若对任意的x∈(0,),f(8(x)

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    B.k∈-00,eC.当k<0时,x+x2<1D若f(x)=k的根记为x,:,g(x)=k的根记为x,x4,且x<<0,w>0,x∈[0,8])的图像,且图像的最高点为S(6,4V3),绿道的后一段为折线段MNP,且∠MNP=120°(如图所示).y0(6.0)(8.0)P(16,0)x(1)求实数A和ω的值以及M、P两点之间的距离;(2)求△MNP面积的最大值18.(本小题满分12分)如图,在棱长为A的正方体ABCD-A,B,C,D,中,M,N分别是AA,CD中点,过D,M,N三点的面与正方体的下底面AB,C,D,相交于直线1.

  • 炎德英才大联考2024年普通高等学校招生考试考前演练四英语试题

    讲三步一体高效训练记Marggraf as director of the Berlin Academy of Sciences.Instead,shying from politics,he chose sim-ple surroundings so that he could work without jealousy or evil intention in his own pharmacy labo-ratory.He was born in Stralsund,Germany,at that time Swedish Pomerania.His parentage wasGerman,and throughout his life he used this language to write down his laboratory notes.He re-ceived an excellent education and experienced an enjoyable childhood painting and collecting insectsand plants.At home he was particularly fascinated with the conversations of the friends of his par-ents,who included doctors and pharmacists.He fell in love with pharmacy books.From borrowedtextbooks of chemistry and pharmacology he devoted himself to absorbing all possible knowledge onthe subjects.He performed over 13,000 experiments in his lifetime,but only a few pieces of laboratory notesand correspondence exist.With his incredible memory Scheele committed virtually everything tomemory and he wrote little.He drove himself constantly,laboring every evening in his laboratory af-ter performing his daily work as a pharmacist.Sir Humphry Davy said of him,"Nothing could dampdown the passion of his mind or chill the fire of his genius:with very small means he accomplishedvery great things..”11.What does the first paragraph mainly tell us about Scheele?A.His short career as a pharmacist.B.His devotion to his work.C.His poor working conditions.D.His achievements in chemistry12.What can be inferred about Scheele?A.He was good at photographing.B.He was admired by his colleagues.C.He cared little about personal fame.D.He preferred to work alone in the lab.13.How did Scheele perfect his knowledge of pharmacy?A.His parents taught him a lot at homeB.He attended a famous medical school.C.He learned from recorded conversations.D.He taught himself through relevant books.14.Humphry Davy's comment is mainly about Scheele'sA.unusual way to conduct experimentsB.extraordinary memory for notesC.passion for and devotion to chemistryD.daily routine working in the lab【答案与解析】本文介绍了瑞典化学家和药剂师Carl Wilhelm Scheele在化学及药剂学方面的研究及成就。ll.D通读第l段可知,本段主要介绍了Carl Wilhelm Scheele在化学研究方面所取得的成就。12.C根据第2段最后两句可知,他曾经拒绝了担任柏林科学院院长的提议,并选择了简单的环境以便继续他的研究。因此,我们可以推断出Carl Wilhelm Scheele不太在乎个人名誉。13.D根据第3段最后两句可知,他爱上了药学书籍,专心致志地从借来的化学和药理学课本中自学与这些学科有关的知识。l4.C根据最后一句可知,Humphry Davy这样评价Carl Wilhelm Scheele:“没有什么能浇灭他心中的激情,也没有什么能熄灭他在化学上的天才之火:他用非常简陋的设备做出了非常伟大的贡献。”因此,Humphry Davy评价了CarlWilhelm Scheele对化学的热爱和奉献。【24新教材,ZCYK,英语-BSD-选择性必修第二册-Y】

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    则下列结论正确的是()n2+n+2A.a,=nn∈N)B.a0=5C.a,a4,as成等比数列D.4+42++am=411.下列四个命题中,错误的是()A.“m≤1"是“关于x的方程m2+2x+1=0有两个实数解的必要不充分条件B.命题“x∈R,使得+x+1<0"的否定是:“对x廷R,均有x+x+1>0”y=Vx2+2+1C.若x>0,则函数√2+2的最小值是2D.若函数.f()=x+3ax+br+a在x=-1有极值0,则a=2,b=9或a=1,b=3.12.已知¥,名分别是函数f)=e+x-2和8()=lnx+x-2的零点,则()A.+X2=2ex+Inx2 =22ve2D.+好<3三、填空题:本题共4小题,每小题5分,共20分.13.数列{a}中,4=2,a1=20,n∈V*.若其前k项和为126,则k=cosπ,00时,f(x)>0,f(k2)+f(41-8-2)>0对任意X∈[1,2]恒成立,则实数k的取值范围是2216.函数f()=r-ax血r在e上不单调,则实数a的取值范围是试卷第3页,共6页

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    5.已知偶数f(x)在(0,+∞)是增函数,则下列结论正确的是小,小1升国2023~2024学年第一学期高三年级期中学业诊断代0的A.f(x)=cosx数学试卷C.f(a)=tD.f(x)=-1(考试时间:上午8:00一10:00)说明:本试卷为闭卷笔答,答题时间120分钟,满分150分。6.几何定理:以任意三角形的三条边为边,向处构造三个等边三角形,则这三个等边三角形四总分的外接圆圆心恰为另一个等边三角形(称为拿破仑三角形)的顶点.在△ABC中,已知题号三171819202122A:w,AC=》3,C名V5,现以边A,BC,C4向外作三个等边三角形,其外接圆得分圆心依次记为D,E,F,则DE的长为B.2V7一、选择题(本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符A.合题目要求的)D.2v5么:1m-1131.已知集合A={xx2-2x<0,B={xlx>1},则A∩B=7.已知f(x)=a(a>0,且a≠1),m,∈(0,+则下列结论正确的是许232'313=2+23:2+8A.(0,+∞)B.(0,2)mn)-f(m)f(n)B.f(m+n)=f(m)+f(n)C.(0,1)D1,2)c"")2fm)22∠2.已知复数x满足(1+i)z=2,则z=22+iB.1-i8.在解决问题"已知1am80=m,请用m表示an20的值”时,甲的结果为m-Y3,乙的结果C.-1+iD.-1-im1+V3m为,则下别结论正确的是2X23.“x2>1”是“x>1”的A.充要条件B.充分不必要条件m2月2mA.甲、乙的结果都正确m2-11+5m¢.必要不充分条件D.既不充分也不必要条件B.甲的结果正确、乙的结果错误-)1片万m加还4.已知a=(1,0),b=(1,1),若(Aa-b)⊥b,则实数入=+5)B.2C.甲的结果错误、乙的结果正确(1+5)A.-2C.-1D.1D、乙的结果都错误4Em-mt百1乃A-+5高三数学第1页(共10页)高三数学第2页(共10页)数

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    2023~2024学年上学期期中质量检测八年级数学注意事项:1.本试卷分试题卷和答题卡两部分,试题卷共4页,三个大题,满分120分,考试时间100分钟.2.试题卷上不要答题,请用0.5毫米黑色签字水笔直接把答案写在答题卡上,答在试题卷上的答案无效。:3.答题前,考生务必将本人姓名、准考证号填写在答题卡第一面的指定位置,一、选择题(每小题3分,共30分)下列各小题均有四个答案,其中只有一个是正确的.1.下列四组数中,属于勾股数的是A.0.3,0.4,0.5B.8,15,17C.2,8,10D.1,2,32在v30,√公,V厄中最简二次根式的个数是A.3个B.2个C.1个D.0个3.如图是小唯关于诗歌《望洞庭》的书法展示,若“湖”的位置345678910望洞庭用有序数对(2,3)表示,那么“螺”的位置可以表示为湖光秋月两相和A.(5,9)潭面无风镜未磨4遥望洞庭山水翠舒B.(9,5)白银盘里仁情螺C.(5,8):D.(8,5)4.一次函数y=-3x+1的图象过点(1y),(x1+1,y2),(1+2,y3),则A.y1

  • 炎德文化数学2024年普通高等学校招生全国统一考试考前演练一答案

    8.如今中国被誉为基建狂魔,可谓是逢山开路,遇水架桥公路里程、高铁里程双双都是世界第一.建设过程中研制出用于基建的大型龙门吊、衡盾构机等国之重器更是世界领先.如图是某重器上一零件结构模型,中间最大球为正四面体ABCD的内切球,中等球与最大球和正四面体三个面均相切,最小球与中等球和正四面体三个面均相切,已知正四面体ABCD棱长为2√6,则模型中九个球的表面积和为()31πA.6πB.9πD.21元4【答案】B【解析】作出辅助线,先求出正四面体的内切球半径,再利用三个球的半径之间的关系得到另外两个球的半径,得到答案如图,取BC的中点E,连接DE,AE,则CE=BE=V6,AE=DE=√24-6=3√2,过点A作AF⊥底面BCD,垂足在DE上,且DF=2EF,所以DF=2√2,EF=√2,故AF=√AD2-DF2=√24-8=4,点O为最大球的球心,连接DO并延长,交AE于点M,则DM⊥AE,设最大球的半径为R,则OF=OM=R,因为Rt△AOM∽R△AEF,所以4O_OMAE EF4-R R即322,解得R=1,MO即OM=OF=1,则AO=4-1=3,故Bsin∠EAF=OM 1A03设最小球的球心为J,中间球的球心为K,则两球均与直线AE相切,设切点分别为H,G,连接HJ,KG,则HJ,KG分别为最小球和中间球的半径,长度分别设为a,b,则AJ=3HJ=3a,AK=3GK=3b,则JK=AK-AJ=3b-3a,高二数学学科参考答案第5页(共20页)

  • 炎德英才大联考2024年普通高等学校招生考试考前演练四英语试题

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  • 炎德英才大联考2024年普通高等学校招生考试考前演练四英语试题

    the holiday in the UK. Yours, Li Hua

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