廣東省三水八年級(jí)上學(xué)期學(xué)業(yè)水平考試數(shù)學(xué)試卷有答案

廣東省三水市八級(jí)學(xué)學(xué)水平考試數(shù)學(xué)試卷說明：．卷共4頁，試用時(shí)分鐘，滿分為120分．2．答卷前，考生務(wù)必用黑色字的簽字筆或鋼筆在答題卡填寫自己的準(zhǔn)考證號(hào)、姓名、試室號(hào)、座位號(hào)．用2B鉛把對(duì)應(yīng)該號(hào)碼的標(biāo)號(hào)涂黑．3題小題選出答案后2B鉛筆把答題卡上對(duì)應(yīng)題目選項(xiàng)的答案信息點(diǎn)涂黑，如需改動(dòng)，用橡皮擦干凈后，再選涂其他答案，答案不能答在試題上．4．非選擇題必須用黑色字跡鋼或簽字筆作答，答案必須寫在答題卡各題目指定區(qū)域內(nèi)相應(yīng)位置上；如需改動(dòng)，先劃掉原來的答案，然后再寫上新的答案；不準(zhǔn)使用鉛筆和涂改液．不按以上要求作答的答案無效．5．考生務(wù)必保持答題卡的整潔考試結(jié)束時(shí)，將試卷和答題卡一并交回．一、選擇題（本大題10小題每小題3分共分）在每小題列出的四個(gè)選項(xiàng)中只有一個(gè)是正確的，請(qǐng)把答題卡上對(duì)應(yīng)目所選的選項(xiàng)涂黑．1．的絕值是A．-6B．6C．±D．2．下列各圖中，經(jīng)過折疊能圍一個(gè)立方體的是A．B．．D．3．中國航母遼寧艦（如題3圖）中國人民海軍第一艘可以搭載固定翼飛機(jī)的航空母艦，滿載排水量為67500噸，這個(gè)數(shù)據(jù)用科學(xué)記數(shù)法表示為

A．6.75×10

B．6.75×10C．6.75×10

噸D．6.75×10

噸4．若()

mn

的值是

題3圖A．-1B．1C2

D．35．一組數(shù)據(jù)2，4，x，，，的眾數(shù)是2則這組數(shù)據(jù)的平均數(shù)、中位數(shù)分別為A．3，

B．3.5，C．3，3.5

D．4，6．如題6圖直線l∥l，∠1=140°∠2=70°則3的度數(shù)是A．60°C．70°

B．65°D．80°7．如果ab，那么下列判斷正確的是A．a(chǎn)＜，b＜0B．a(chǎn)＞0，＞C．a(chǎn)≥，b≤0D．a(chǎn)＜0，＞或＞，b＜8．不等式組的集數(shù)軸上表示正確的是x

題6圖

＝＝ABCD9．下列標(biāo)志中，可以看作是中對(duì)稱圖形的是ABCD10．次函數(shù)y

的圖象如題圖示，那么一函數(shù)

y

與反比例函數(shù)

在同一平面直角坐標(biāo)系中的大致圖象為題10圖ABCD二、填空題（本大題6小題，每題4分共24）請(qǐng)將下列各題的正確答案填寫在答題卡相應(yīng)的位置上．11．解因式：2x

x

．12．知

，則

__________．13．∠=80°，則α的角為__________．14．題14圖在山的東側(cè)點(diǎn)一個(gè)熱氣球，由于受西風(fēng)的影響，以30米分速度沿與地面成75°角的方向飛行25分鐘后到達(dá)C處此時(shí)熱氣球上的人測(cè)得小山西側(cè)B點(diǎn)的角為30°，則小山東西兩側(cè)、兩間的距離__________．15．題15圖，在△中，∥BCAD=1AB，=2，則BC=__________．16．題16圖，AB是半圓O直徑，且AB，C為圓上的點(diǎn)．將此半圓沿所在的直線折疊，若圓弧恰過圓心，則圖中陰影部分的面積（果保留π題14圖

題圖

題圖三、解答題（一本大題3小，每小題分，15分17．方程：

52(8

．18．化簡(jiǎn)，再求值：

3xx2()x

，再選擇一個(gè)使原式有意義的代求值．19．題19圖，已知線段AB

（）出線段AB的直分線l用尺規(guī)作圖，保留作圖痕跡，不要求寫出作法（）（1）中所作的直線l上意取兩點(diǎn)，（線段AB的上方結(jié)AM，ANBM，BN．求證：∠=∠MBN題19圖四、解答題（二本大題3小，每小題分，24分20．商場(chǎng)以每件280元價(jià)購進(jìn)一批商品，當(dāng)每件商品售價(jià)為360元，每月可售出60件為了擴(kuò)大銷售，商場(chǎng)決定采取適當(dāng)降價(jià)的方式促銷，經(jīng)調(diào)查發(fā)現(xiàn)，如果每件商品降價(jià)1元那么商場(chǎng)每月就可以多售出件（）價(jià)前商場(chǎng)每月銷售該商品的利潤是多少元？（）使商場(chǎng)每月銷售這種商品的利潤達(dá)到7200元，且更有利于減少庫存，則每件商品應(yīng)降價(jià)多少元？21．障房建設(shè)是民心工程，廣省某市從年開始加快保障房建設(shè)進(jìn)程，現(xiàn)統(tǒng)計(jì)了該市2009年2013年年新保障房情況，繪制成如題22圖示的折線統(tǒng)計(jì)圖和不完整的條形統(tǒng)計(jì)圖．某市2009～2013年新建保障房數(shù)年增長(zhǎng)率折線統(tǒng)計(jì)圖

某市2009～年新建保障房套數(shù)條形統(tǒng)計(jì)圖20092010201120122013年20092010201120122013年份題22圖（）麗看了統(tǒng)計(jì)圖后說市2012年建保障房的套數(shù)比2011年了認(rèn)小麗說法正確嗎？請(qǐng)說明理由；（）補(bǔ)全條形統(tǒng)計(jì)圖；（）這5年均每年新建保障房的套數(shù)．22．題22圖．點(diǎn)A、、、⊙上，⊥于點(diǎn)，過點(diǎn)作OFBC于F求證）△AEB△OFC；（）=2FO．題22圖五、解答題（三本大題3小，每小題分，27分

23．題23圖，拋物線2（a＞0）經(jīng)過原點(diǎn)O和點(diǎn)A（，（）接寫出拋物線的對(duì)稱軸與軸的交點(diǎn)坐標(biāo)；（）（x，y，y）拋物線上，若＜＜，比較y的?。唬ǎ〣-1，）該拋物線上，點(diǎn)C與B關(guān)拋物線的對(duì)軸對(duì)稱，求直線的函數(shù)關(guān)系式．題23圖24．題24圖直線直平分⊙的徑OA于點(diǎn)B，交⊙O于點(diǎn)、，是⊙的線為點(diǎn)，連結(jié)AE，交于點(diǎn)F（）⊙的徑為8，的長(zhǎng)；（）證PE；（）PF=13，A=，的長(zhǎng)．題24圖25．知：如圖①，在平行四邊中，=6AD．AD為斜邊在平行四邊形ABCD的內(nèi)部作eq\o\ac(△,Rt)，∠=30°∠°．（）△的長(zhǎng)；（2）△每秒2個(gè)單位長(zhǎng)度的速度沿向平行移動(dòng)，得到△D，AD與重合時(shí)停止移動(dòng)運(yùn)動(dòng)時(shí)間為t秒eq\o\ac(△,，)ED與△重的面積為，請(qǐng)直接寫出與t之間函數(shù)關(guān)系式，并寫出t的取范圍；（）圖②，在），當(dāng)停止移動(dòng)后得到△，BEC繞點(diǎn)按時(shí)針方向旋轉(zhuǎn)（°＜α＜180°旋過程中的應(yīng)點(diǎn)為B，的應(yīng)點(diǎn)為E，設(shè)直線B與線交于點(diǎn)P與線CB交于Q是否存在這樣的使BPQ為等腰三角形？若存在，求出度數(shù)；若不存在，請(qǐng)說明理由．題25圖

題25圖

參考答案及評(píng)分標(biāo)準(zhǔn)數(shù)

學(xué)一、選擇題（本大題小題每小題分共分）題號(hào)答案

B

A

A

D

B

C

D

D

D

B二、填空題（本大題小題，每小題4分，共分）．x(x

．

．14750

．

．

三、解答題（一本大題3小題，小題分，共15分17解：x+2=8+··························································································x=6······························································································x=3．···························································································18解：原=

2(x(x(x2)(2)

····························································=2+8·························································································3分當(dāng)x=1時(shí)原式=2×1+8=10的值不能為2或）······························5分19）：如圖所示：················································································1分·······················································2分（2證明：∵l是垂平分線∴=BMAN=BN，·································································∴∠=∠MBA，∠NAB∠NBA·············································4分∴∠∠NAB∠MBA∠NBA，即：∠MAN=∠．································································四、解答題（二本大題3小題，小題分，共24分20解）題意，得60-280）元（2設(shè)要使商場(chǎng)每月銷售這種商品的利潤達(dá)到7200，且更有利于減少庫存，則每件商品應(yīng)降價(jià)元，由題意，得（360-x+60），解得：x，x12∵有利于減少庫存，∴x=60．答）價(jià)商場(chǎng)每月銷售該商品的利潤是4800元）使商場(chǎng)每月銷售這種商品的利潤達(dá)到元且更有利于減少庫存，則每件商品應(yīng)降價(jià)60．

AE，∴AE，∴21解）市2012年建保障房的增長(zhǎng)率比的增長(zhǎng)率減少了，但是保障房的總數(shù)在增加，故小麗的說法錯(cuò)誤．···················································（2年障房的套數(shù)為：750×(1+20%)=900套·····························3分年保障房的套數(shù)為：x(1+20%)=600，則x=500···························4分如圖所示：··········································6分（3（）·············································答：這年平均每年新建保障房的套數(shù)為784套······························8分22證明）圖，連接，則=∠，··········································1分∵⊥，∴∠∠，∴∠BAE=∠COF，······················又∵AC，⊥，∴∠∠AEB，···························3分∴△∽△；···································································（2∵∽△，∴，················································BEFC由圓周角定理，=∠，∠DAE=∠CBE∴△∽，∴

AEFOADBC

，····························∵⊥，∴BC，∴=FOFO，即FO············································································8分五、解答題（三本大題3小題，小題分，共27分23解）物線的對(duì)稱軸與軸的交點(diǎn)坐標(biāo)為，0···································（2拋物線的對(duì)稱軸是直線x．由圖可知，當(dāng)x＜1時(shí)隨的增大而減小，····································

kk3∴當(dāng)x＜＜1時(shí)y＞；····························································12（3∵對(duì)稱軸是x=1，（1）在該拋物線上，點(diǎn)與關(guān)于拋物線的對(duì)稱軸對(duì)稱，∴點(diǎn)C坐標(biāo)是，····························································設(shè)直線的系式為=kx+b（k≠0，·························6分解得．··············································································∴直線的數(shù)關(guān)系式是：=24．··············································24解）接OD，··················································································∵直線PD垂直平分O的徑于B⊙O的徑為8，∴OB=4，BC=BDCD，························································∴在eq\o\ac(△,Rt)OBD中BDOD

，∴=3

；········································································（2∵PE⊙O的線∴PEO，···············································∴∠PEF-∠，=∠-∠，································∵OE，∴A∠AEO∴∠PEF=∠，∴PE；·························································6分（3過點(diǎn)P作PG⊥于，····························································∴∠==90°，∵∠=，∴∠∠，∴=A=13×···························································∵PE=PF∴=10．····························································9分25解）△的長(zhǎng)為3；····························································1分（2①S＝t2（0t································································2

33