2024届南宁市初中毕业班适应性测试数学参考答案一、选择题（本大题共12小题，每小题3分，共36分）题号123456789101112答案ABACABDACBDC二、填空题（本大题共6小题，每小题2分，共12分）113．x(x-5)；14．x≥3；15．；16．（0，6）；17．25；18．nm2．3三、解答题（本大题共72分）19．（本题满分6分）解：原式＝9（1）+2，·····························3分＝-9+2，·····································4分＝-7．········································6分20．（本题满分6分）解：原式＝a22abb22abb2，·············2分＝a24ab．································4分1当a＝2，b＝-时，41原式＝2242（），·····················5分4＝4-2＝2．··········································6分21．（本题满分10分）（1）如图所示，CE即为所求；·····················5分（2）证明：∵在Rt△ABC中，∠ACB=90°，∠A=40°，∴∠B=90°-40°=50°．··················6分又点D为AB中点，∴CD=BD．·······························7分（第21题图）∴∠B=∠DCB=50°．·················8分∵CE平分∠BCD，1∴∠BCE=∠BCD=25°．···········9分2∴∠AEC=∠B+∠BCE=75°．········10分数学参考答案第1页（共4页）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGhFAEIAAByAFABAA=}#}22．（本题满分10分）（1）m＝40，a＝12，n%＝40%；·················3分（2）他的说法是错误的．····························4分理由如下：∵在参加测试的40名学生测试成绩中，排在最中间的两个分数都是85，8585∴中位数为85．·····················6分2∵84＜85，∴有一半以上的同学成绩超过了84分．···7分∴小邕的说法是错误的．（3）解：500×(40%+30%)＝350（人）············9分答：估计本年级中食品安全意识良好的学生人数为350人．····················10分23．（本题满分10分）（1）证明：∵AB∥DE，∴∠B＝∠E．······································1分又BF＝CE，∴BF+FC＝CE+FC，即BC=EF．·············2分在△ABC和△DEF中ABDEBE，·····································3分BCEF∴△ABC≌△DEF（SAS）．···················4分（2）解：连接AD交FC于点O，∵△ABC≌△DEF，∴AC＝DF，∠ACB＝∠DFE．················5分∴AC∥DF．∴四边形ACDF是平行四边形．（第23题图）又AF＝FD，∴四边形ACDF是菱形．·······················6分1∴AD⊥CF，OF＝OC＝FC＝2．2在Rt△ACO中∴AOAC2OC2(13)2223，··7分∴AD＝2AO＝6．··································8分11∴S四边形ACDF=FCAD6412．·····10分22数学参考答案第2页（共4页）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGhFAEIAAByAFABAA=}#}24．（本题满分10分）（1）解：设团购群1《儒林外史》和《简爱》的单价分别是x元、y元；············1分3x2y12220由题意得：，·····3分4x3y288x48解得．···························5分y32答：团购群1《儒林外史》和《简爱》的单价分别是48元、32元．············6分（2）解：团购群1：（4832）150.784（0元），·······································7分团购群2：7015105（0元）1050-40393（0元），····8分∵840＜930，·································9分∴选择团购群1购买更合算．············10分25．（本题满分10分）（1）证明：如图1，连接OC，·····················1分∵OA＝OB，CA＝CB，∴OC⊥AB.····································3分又OC为半径，∴AB是⊙O的切线.························4分（2）解：设半径为R，1在Rt△OCB中，∠OCB=90°，BC＝AB＝4，（第25题图1）22OB2OC2BC2，即R2R242，解得R＝3，···································5分∴OB＝OE+BE＝5.··························6分方法一：如图2，过O作OM⊥EG于点M，又OA＝OB，AC＝BC，∴∠A＝∠B.又OF＝OE，∴∠F＝∠OEF＝∠BEG.∴△AFG∽△BEG.··························7分∴∠AGF＝∠BGE＝90°，（第25题图2）FGAGAF84.EGBGBE2∴FG＝4EG，AG＝4GB.∴AB＝4GB+GB＝8.8解得GB=.··································8分5222286∴EGBEBG2.·············································9分55618∴EFFGEG3EG3.················································10分55数学参考答案第3页（共4页）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGhFAEIAAByAFABAA=}#}方法二：如图3，连接DE，ODOE3∵∠DOE＝∠AOB，，OAOB5∴△DOE∽△AOB.·························7分DEOE3∴.ABOB524∴DE=.····································8分（第25题图3）5∵DF是⊙O直径，∴∠DEF＝90°.·····························9分22222418∴EFDFDE6.··········································10分5526．（本题满分10分）（1）解：由题意可知，抛物线顶点为（2，2），经过B（0，1.2）2设抛物线的解析式为yax22···············································1分2将点B（0，1.2）代入，得1.2a022，1解得a.·············································································2分512∴抛物线的解析式为yx22.············································3分5（2）队伍排列时，最高的队员在正中间其位置的横坐标为2，其余同学按照从高到低的顺序在其两侧对称排列.身高为1.68与1.73的队员所在位置的横坐标分别为1.5与2.5，···············4分身高为1.60的队员所在位置的横坐标分别为1与3；∵1.8＜2，∴身高最高的队员可以通过.·····························································5分1212由题意，把x1.5代入yx22中，得y1.5221.95.55∵1.95＞1.73＞1.68，∴可以通过.··················································································6分1212由题意，把x1代入yx22中，得y1221.8.55∵1.8＞1.60，∴可以通过.··················································································7分综上所述，所有队员都可以通过.12（3）由题意，把y1.60代入yx22中，512则1.6x22.·····································································8分5解得＋（舍去）.分x122,x122···················································9∵22＞0.5，∴最左边的跳绳队员与离他最近的甩绳队员之间距离的取值范围：22＜x≤1.··············································································10分数学参考答案第4页（共4页）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGhFAEIAAByAFABAA=}#}