2021北京平谷區(qū)初一上期末數(shù)學(xué)試卷及答案

2021北京平谷初一（上）期末數(shù)學(xué)2021.1（考試時(shí)間120分鐘100分）學(xué)校____________________________________________1．本試卷共6頁(yè)，共三道大題，24道小題．考生須知2．在試卷和答題卡上認(rèn)真填寫(xiě)學(xué)校、班級(jí)、姓名、考號(hào)．3．試題答案一律填涂或書(shū)寫(xiě)在答題卡上，在試卷上作答無(wú)效．4．在答題卡上，選擇題、作圖題用鉛筆作答，其他試題用黑色字跡簽字筆作答．5．考試結(jié)束，請(qǐng)將本試卷、答題卡和草稿紙一并交回．一、選擇題（本題共24分，每小題3分）下面1-8題均有四個(gè)選項(xiàng)，其中符合題意的選項(xiàng)只有一個(gè)．．．1.12月3日23點(diǎn)10分，嫦娥五號(hào)上升器月面點(diǎn)火，約6分鐘后，順利將攜帶月壤的上升器送入預(yù)定環(huán)月軌道，實(shí)現(xiàn)我國(guó)首次地外天體起飛.起飛前，國(guó)旗展示系統(tǒng)成功在月面打開(kāi)，這是中國(guó)首次在月球展示“織物版五星紅旗.380000公里外，那一抹“中國(guó)紅振奮著每一個(gè)中國(guó)人的心.請(qǐng)你用科學(xué)記數(shù)法表示380000:A.384B.3.8105C.3.8106D.0.381062．下列各組數(shù)中，互為相反數(shù)的一組是A.?(和3B.(?3和-33C.?3和?3D.(2和?323．如圖是一個(gè)小正方體展開(kāi)圖，把展開(kāi)圖折疊成小正方體后，有“志字一面的相對(duì)面上的字是A.事竟4.下列等式變形正確的是x=2C.成D.者A．若2x1，則=B．若4x?1=2?3x，則4x+3x=2?132x=3，則x=?C.若21/93x+11?2xD．若?=1，3(3x+??2x)=1235．下列各式是同類(lèi)項(xiàng)的是A.2bB.2a2bab2C.2aD.2和26．有理數(shù)b在數(shù)軸上的位置如圖所示，下列說(shuō)法正確的是A.0B.abC.a+b0D.?ab7．下列語(yǔ)句正確的個(gè)數(shù)是①直線外一點(diǎn)與直線上各點(diǎn)連接的所有線段中，垂線段最短②兩點(diǎn)之間直線最短③在同一平面內(nèi)，兩條不重合的直線位置關(guān)系不平行必相交④兩點(diǎn)確定一條直線A．1B．2C．3D．48.大家都知道，七點(diǎn)五十可以說(shuō)成差十分鐘八點(diǎn)，有時(shí)這樣表達(dá)更清楚，這也啟發(fā)了人們?cè)O(shè)計(jì)了一種新的加減記數(shù)法．——比如：8寫(xiě)成12，12=10-2—189229=200-20+9———7683寫(xiě)成12323=10000-2320+3————按這個(gè)方法請(qǐng)計(jì)算5231-3241=A.2408B.1990C.2410D.3024二、填空題（本題共24分，每小題3分）9.北京某天最高氣溫是6，最低氣溫是-3℃，則這一天的溫差是____________ab10.請(qǐng)你寫(xiě)出一個(gè)只含有,，且系數(shù)為，次數(shù)為3的單項(xiàng)式是________14161211.計(jì)算：(+?)=________12.已知a4(b3)++?2=0,則(a+b2021=________13.計(jì)算：60?=________2/914.《九章算術(shù)》中有一道闡述“盈不足術(shù)”的問(wèn)題，原文如下：今有人共買(mǎi)物，人出八，盈三；人出七，不足四．問(wèn)人數(shù)，物價(jià)各幾何？譯文為：現(xiàn)有一些人共同買(mǎi)一個(gè)物品，每人出8元，還盈余3元；每人出7元，則還差4元，問(wèn)共有多少人？這個(gè)物品的價(jià)格是多少？設(shè)這個(gè)物品的價(jià)格是x元，則可列方程為15．如圖，C是線段AB上一點(diǎn)，D，E分別是線段AC，BC的中點(diǎn)，若AC=2，BE=4，則DE=．ADCEB16．觀察右面的圖案，每條邊上有（n≥2）個(gè)方點(diǎn),每個(gè)圖案中方點(diǎn)的總數(shù)是S.（1）請(qǐng)寫(xiě)出n=5時(shí)，S=（2）按上述規(guī)律，寫(xiě)出S與n的關(guān)系式，S=；．三、解答題(19-20題每題分，21-24題每題525題626題652分)17．計(jì)算：（1）2?(+(14)?(?12)13(+(7???22（2）2418．解方程：（1）x?6)=?4x?32x+11?10x=1?（2）3619．按要求畫(huà)圖，并回答問(wèn)題：如圖，在同一平面內(nèi)有三點(diǎn)A、B、C．（1）畫(huà)射線AC和直線AB；（2）連接線段BC，并延長(zhǎng)BC至D，使CD=BC；（3）連接線段AD；（4）通過(guò)畫(huà)圖和測(cè)量，點(diǎn)C到線段AD的距離大約是（精確到0cm）．3/9(?2+2x+2x)+1，其中x=3)(2x?3x2220．先化簡(jiǎn)，再求值：21．已知：是∠AOC的角平分線，OC是∠AOD的角平分線，∠COD=40°．分別求∠AOD和∠BOC的度數(shù)．DCOBA222020年的天貓雙十一比以往來(lái)的更早一些.如今的雙十一也不再是當(dāng)年那個(gè)僅此一天的雙十一,今年的活動(dòng)期已經(jīng)拉長(zhǎng)到了一個(gè)月左右.曉晨一家人打算在今年的雙十一促銷(xiāo)中，爭(zhēng)取花最少的錢(qián)，買(mǎi)到物美價(jià)廉的產(chǎn)品.曉晨想買(mǎi)一些學(xué)習(xí)用品，清單見(jiàn)表1.媽媽想買(mǎi)一臺(tái)智能掃地機(jī)器人，爸爸想買(mǎi)一臺(tái)空氣凈化器，經(jīng)過(guò)反復(fù)的篩選，一家人決定從以下兩個(gè)品牌當(dāng)中挑選掃地機(jī)器人和空氣凈化器，它們的單價(jià)見(jiàn)表2.雙十一電子商品促銷(xiāo)方案見(jiàn)表3.表1表2單價(jià)（元）總價(jià)（元）掃地機(jī)器人單價(jià)空氣凈化器單價(jià)（元/臺(tái)）（元/臺(tái)）筆記本碳素筆書(shū)包580甲品牌乙品牌2600300025002400260（1）曉晨購(gòu)買(mǎi)a個(gè)筆記本，b支碳素筆，1個(gè)書(shū)包一共要支付（用含有字母a、b的代數(shù)式來(lái)表示）元.（2）曉晨購(gòu)買(mǎi)筆記本的數(shù)量比購(gòu)買(mǎi)碳素筆的數(shù)量少3個(gè)，還購(gòu)買(mǎi)了一個(gè)書(shū)包，總金額請(qǐng)見(jiàn)表，請(qǐng)問(wèn)曉晨購(gòu)買(mǎi)了幾支碳素筆？（3）請(qǐng)你幫忙計(jì)算選擇哪種品牌的掃地機(jī)器人和空氣凈化器能夠花費(fèi)最低,并直接寫(xiě)出總花費(fèi)為元.4/9表31.所有電子商品均享受每滿(mǎn)300減元；2.在滿(mǎn)減的基礎(chǔ)上還可享受購(gòu)買(mǎi)同一品牌商品一件9折、兩件8折的優(yōu)惠；3.掃地機(jī)器人預(yù)售定金翻倍：提前支付50元定金抵20010月21日-11月日期間支付元定金，可在11月11日結(jié)算時(shí)抵扣200元）23．如圖：點(diǎn)O為直線上一點(diǎn)，過(guò)點(diǎn)O作射線OP，使∠AOP=60°，將一直角三角板的直角頂角放在點(diǎn)O處（1）如圖，一邊OM為射線OB上，另一邊ON在直線AB的下方，那么鈍角∠PON的度數(shù)（2）如圖，將圖1中三角板繞點(diǎn)O逆時(shí)針旋轉(zhuǎn)，使邊OM在∠BOP的內(nèi)部，且OM恰好平分∠BOP，此時(shí)∠BON的度數(shù)（3）如圖，繼續(xù)將圖2中的三角板繞點(diǎn)O逆時(shí)針旋轉(zhuǎn)度，使得ON在∠AOP內(nèi)部，且滿(mǎn)足∠AOM=3∠NOP時(shí)，求的度數(shù)26.閱讀下面材料，回答問(wèn)題已知點(diǎn)A,B在數(shù)軸上分別表示有理數(shù)a，b.A,B兩點(diǎn)之間的距離表示AB（一）當(dāng)AB兩點(diǎn)中有一點(diǎn)在原點(diǎn)時(shí)，不妨設(shè)點(diǎn)A在原點(diǎn)，如圖1)BO==b?a=b?a=a?b（二）當(dāng)A,B兩點(diǎn)都不在原點(diǎn)時(shí),①如圖2,點(diǎn)A,B都在原點(diǎn)的右邊,5/9ABO=?=b?a=b?a=a?b②如圖3,點(diǎn)A,B都在原點(diǎn)的左邊,BAO=?=b?a=?b?(?a)=a?b=a?b③如圖4,點(diǎn)A,B在原點(diǎn)的兩邊,BAO=+=a+b=a+(b)=a?b=a?b綜上,數(shù)軸AB兩點(diǎn)的距離=a?b利用上述結(jié)論,回答以下幾個(gè)問(wèn)題:（1）數(shù)軸上點(diǎn)A表示的數(shù)是1，點(diǎn)B表示的數(shù)是，且點(diǎn)B與點(diǎn)A在原點(diǎn)的同側(cè)，AB=3，則x=__________（2）數(shù)軸上點(diǎn)A到原點(diǎn)的距離是1B表示的數(shù)絕對(duì)值是，則AB=_________（3）若點(diǎn)、B在數(shù)軸上表示的數(shù)分別是-4、2P在數(shù)軸上表示的數(shù)是x，當(dāng)8時(shí)，直接寫(xiě)x+=的值6/92021北京平谷初一（上）期末數(shù)學(xué)參考答案一、選擇題（本題共24分，每小題3分）題號(hào)答案12345678BDACCDCA二、填空題（本題共24分，每小題3分）題號(hào)答案991011-112-11347°22’14151616，4n?42a2b或2ab28x?3=7x4+5三、解答題(17-18題每題分，19-22題每題523題624題652分)171）．解：2?(+(14)?(?12)2?1?+=···················································································4分=5···································································································5分13(+(7???22172）．解：241=1+3?2······················································································3分2=2+3?2···························································································4分=3····································································································5分181）．解：x?6)=?4x?3去括號(hào)，得5x?=4x?3·······························································2分移項(xiàng)，得5x+4x=3+···································································3分合并同類(lèi)項(xiàng)，得9x·······································································4分=系數(shù)化為1x=3············································································5分2x+11?10x182）.解：=1?36去分母（方程兩邊同乘以6），得2(2x+=6??10x)．····································································1分4x+2=6?1+10x去括號(hào)，得．·························································2分7/9移項(xiàng)，得4x?10x=6?1?2．···························································3分6x=3．···································································4分合并同類(lèi)項(xiàng)，得1系數(shù)化為1x=?．······································································5分219.圖略·································································································4分距離在1.41.6之間··············································································5分()2x?3x2?2+2(x+2x2)+120.解：2x?3x4x+x當(dāng)x=3時(shí)，2+2+2x+4x2+1·································································2分＝＝2+3．····················································································3分?34+(?+3··········································································4分2原式==05分平分AOD,AOC=COD=AOD21.解：1…………………….………1分2又=40==40平分…………………….………2…………………….………3分分1…………………….………4分=2=…………………….………5分22．解：（）ab++…………………….………1分（2）設(shè)曉晨購(gòu)買(mǎi)了x支碳素筆根據(jù)題意列方程，得……2分2x+x?+60=80.………….………3分……………4分解得：x=5答：曉晨購(gòu)買(mǎi)了5支碳素筆。（3）3386…………………….………5分23.解：（）NOP=150…………………….………1分8/9（2）+=180=………………….………2分=平分BOP1…………………….………3分BOM=BOP=602=90……………….………4分BON=