哈爾濱市南崗區(qū)九年級(jí)數(shù)學(xué)期末試題及答案哈爾濱市南崗區(qū)九年級(jí)數(shù)學(xué)期末試題及答案哈爾濱市南崗區(qū)九年級(jí)數(shù)學(xué)期末試題及答案2021—2021學(xué)年度〔上〕九年級(jí)調(diào)研測試數(shù)學(xué)試卷參照答案及評(píng)分標(biāo)準(zhǔn)一、〔每小號(hào)12AD

3分，共3D

30分〕45CD

6B

7B

8A

9A

10C二、填空〔每小3分，共30分〕號(hào)1112131415答案1a(ab)(ab)x5413號(hào)1617181920答案10422π2或8①②③④2三、解答〔此中21—22各7分，23—24各8分，25—27各10分，共60分〕21.〔安分7分〕解：a2(12a3)a2(a1)2a3a2a11.?.5′a21a1(a1)(a1)a1(a1)(a1)(a2)a1∵a21∴原式=112????′(21)122.....222.〔安分7分〕.(1)畫正確????????.............′??，A.....(83,3)???.....2′(2)BC=32110.?..........................................................................′1弧BB的是901010.........................................................′1223.〔安分8分〕(1)明：由旋可知，△ABD≌△ACD'???.........................′?.1AD=AD',∠BAD=∠D'AC∴∠BAD+∠DAC=∠D'AC+∠DAC即∠BAC=∠DAD'∵∠BAC=120°，∠DAE=60°∴∠D'AE=∠DAE=60°?..................1′又∵AE=AE∴△ADE≌△AD'E∴DE=D'E....................................1′(2)：∠DAE=1∠BAC...............................................................′12由〔1〕可知，AD=AD'又∵AE=AEDE=D'E∴△ADE≌△AD'E....................1′∴∠D'AE=∠DAE∴∠DAE=1∠DAD'=1∠BAC...............................1′222.............................2′〔3〕2AAD'D'BDEC〔第23題圖1〕〔第23題圖1〕

BCDE〔第23題圖2〕〔第23題圖2〕24.〔安分8分〕1〕ABCD(A,D)(B,D)(C,D)E(A,E)(B,E)(C,E)..........................3′〔2〕由表格能夠看出，從甲、乙兩種品牌中各一種型號(hào)的可能出的果有6種，而且它被中的可能性同樣，A型號(hào)被中〔事件A〕的果有2種，即21.........................................3′〔A,D〕〔A,E〕,因此P(A)36〔3〕購買A型號(hào)電腦的方案有兩種，設(shè)購買A型號(hào)電腦x臺(tái).假定用〔A,E〕方案，那么購買E型號(hào)電腦〔36-x〕臺(tái)，依題意可知6000x+2000〔36-x〕=100000解得x=7....................................1′假定用〔A,D〕方案，那么購買D型號(hào)電腦〔36-x〕臺(tái)，依題意可知6000x+5000〔36-x〕=100000，解得x=-80<0不合題意，舍去..................................................1′答：購買的A型號(hào)電腦有7臺(tái).25.〔本題總分值10分〕〔1〕解：DC//BF............................................................′1在⊙O中，∵AB是直徑，CD是弦，DE=CE∴AB⊥CD∵BF切⊙O于B∴AB⊥BF∴∠AED=∠ABF=90°∴DC//BF...........2′〔2〕∵HG⊥BC∴∠EGC=90°=∠BEC∴∠C+∠CEG=90°∠CEG+∠BEG=90°∴∠BEG=∠C........................................1′∵∠BEG=∠HEA,∠A=∠C∴∠A=∠HEA.................................1′同理可證∠ADE=90°-∠A,∠HED=90°-∠HEA∴∠HDE=∠HED...................1′∴AH=HE=HD,即EH是△ADE的中線..........................1′(3)過點(diǎn)D作BF的垂線，垂足為K.由〔2〕可知，DH=HE=EC=DE∴△DHE為等邊三角形∴∠ADE=60°=∠F∴∠FDK=30°∴FK=1DF922在Rt△DKF中，DK=DF2FK292(9)293..........................1′22∵∠DEB=∠EBK=∠BKD=90°∴四邊形DEBK為矩形∴DK=BE=932∵AB為直徑∴∠ADB=90°∴∠A=∠BDE=90°-60°=30°.....................1′在Rt△DBE中，BD=2BE=93在Rt△ABD中，AB=2BD=183∴OA=93..................................1′26.〔安分10分〕解：〔1〕S2x28x16????????4′〔2〕W120x241x(4x)6080(Sx2)80x2160x1280..............3′2a800∴當(dāng)xb160...................1′2a21，80W有最小4acb24801280(160)21200〔元〕.................2′4a48027.〔安分10分〕解：(1)點(diǎn)A、B的坐標(biāo)分別為〔1，0〕、〔0，3〕.在△AOB中，OA=1，OB=3,∴AB1(3)22...............................................................................′1取斜AB的中點(diǎn)W,接OW,OW=AW=1AB=1=OA∴△AOW等三角形2∴∠A=60°∠ABO=90°-60°=30°在△BEF中，令EF=x，BF=2x由勾股定理得x2(3t)2(2x)2可求x=t∴AD=EF=t∵AD//EF∴四形ADEF平行四形....................................................1′∴SADOEt(33t)3t23t〔0t1〕.′.........................12〕∵∠ADF=90°EF//OD∴∠DFE=∠ADF=90°=∠ODF=∠DOE1∴四邊形ODFE是矩形∴EF=OD∴t1t解得t1′2∴BE=3，點(diǎn)E的坐標(biāo)為〔0，3〕，點(diǎn)G的坐標(biāo)為〔2,3〕...............................1′222設(shè)對(duì)稱軸l與EG的交點(diǎn)為S∵M(jìn)S//BE點(diǎn)E與點(diǎn)G對(duì)于直線l對(duì)稱∴ES=SG點(diǎn)M在斜邊13∴極點(diǎn)M的坐標(biāo)為〔1，33〕...........1′BG上，可得MS=BE=244∴拋物線的分析式為ya(x1)233∵經(jīng)過點(diǎn)E〔0，3〕∴423a(01)233a3所求拋物線的解析式為244y3(x1)233.............................................................................′144即y3x23x3422(3)平移后新拋物線的分析式為y3x233，新拋物線與x軸的交點(diǎn)坐標(biāo)為44R(3,0),Q(3,0).延伸RN至點(diǎn)T，使NT=RN，連結(jié)TQ.又∵OR=OQ1TQ∵四邊形PKNH為平行四邊形∴ON//PKNH=PK∴ON//TQ且ON=2∴TQ//PK∴∠CPK=∠CQT∠CKP=∠CTQ又∵CP=CQ∴△PCK△QCT∴PK=TQ=2ON∴NH=2ON∵點(diǎn)H是新拋物線與y軸的交點(diǎn)∴OH=33∴PK=NH=2OH=3...........................................1342設(shè)知足要求的點(diǎn)P的坐標(biāo)為(p,3p233),那么點(diǎn)K的橫坐標(biāo)為p.設(shè)直線RN的分析式為440k(3)bk14ykxb那么33，解得3b4k0b4∴直線RN的分析式為y1x3