八年级数学答案

资阳市2017-2018学年度学业质量检测八年级数学试题参考答案及评分意见说明：1.解答题中各步骤所标记分数为考生解答到这一步应得的累计分数。2.参考答案一般只给出该题的一种解法，如果考生的解法和参考答案所给解法不同，请参照本答案及评分意见给分。3.考生的解答可以根据具体问题合理省略非关键步骤。4.评卷时要坚持每题评阅到底，当考生的解答在某一步出现错误、影响了后继部分时，如果该步以后的解答未改变问题的内容和难度，可视影响程度决定后面部分的给分，但不得超过后继部分应给分数的一半；如果这一步后面的解答有较严重的错误，就不给分；若是几个相对独立的得分点，其中一处错误不影响其他得分点的得分。5.给分和扣分都以1分为基本单位。6.正式阅卷前应进行试评，在试评中须认真研究参考答案和评分意见，不能随意拔高或降低给分标准，统一标准后须对全部试评的试卷予以复查，以免阅卷前后期评分标准宽严不同。一、选择题（每小题3分，共10个小题，满分30分）1-5.DBCAC；6-10.DBDDA二、填空题：（本大题共6个小题，每小题3分，共18分）11.-1；12.6；13.83；14.12xx；15.4；16.(2016，0)；三、解答题：（本大题共8个小题，共72分）解答应写出必要的文字说明、证明过程或演算步骤.17.原式=11a，·····················································································4分当a=3时，原式=14.················································································8分18.（1）众数为8万车次，中位数为8万车次，平均数为8.5万车次；·················3分（2）30×8.5=255(万车次).答:估计4月份共租车255万车次；···························5分（3）3200×0.75÷9600=25%.答:全年租车费收入占总投入的25%.························8分19.（1）在平行四边形ABCD中，∠BAD的角平分线AE交CD于点F，易证AB=BE，··························································································2分又∵AB=CD，∴BE=CD.·············································································4分（2）由BE=CD=AB，∠BEA=60°得△ABE为等边三角形，则AE=4，···················5分又∵BF⊥AE，∴AF=EF=2，根据勾股定理得BF=23，··································6分易证△ADF≅△ECF，··················································································7分∴平行四边形ABCD的面积等于△ABE的面积，则面积等于43.························8分20.（1）设A种运动鞋的进价为x元，依题意得，3200256020xx，解得x=100，···························································································3分经检验，x=100是原分式方程的解，所以，x=100；·········································4分则A运动鞋的售进价价为100元/双，B运动鞋的进价是80元/双，（2）设总利润为W，则W=（250-100）m+（180-80）（200﹣m）=50m+20000，·································7分∵50＞0，W随m的增大而增大，································································8分又∵90≤m≤105，所以，当m=105时，W有最大值，即此时应购进甲种运动鞋105双，购进乙种运动鞋95双.·····················································································9分21.（1）依题意可知，点A的横坐标为-1，代入23yx，求出A的坐标为（-1，3），则y1的解析式为y=-x+2；·····································3分（2）∵y=2x+b与x轴交于点B（3，0），·····················································4分则直线BC的解析式为y=2x-6，···································································6分求出点C的坐标为（83，23），·································································7分∴SAOC=18(1)223=113.············································································9分22.（1）易证BM=MD=DN；·······································································2分∴四边形BMBN为菱形；··········································································4分（2）设BM=x，在Rt△AMB′中，利用勾股定理求出x=133，·······························5分则DM=133=DN，······················································································6分过点M作MQ⊥CD于点Q，则NQ=1313(6)33=83，······································7分在Rt△MNQ中，利用勾股定理可得MN=4133.·············································9分23.（1）易求点A的坐标为（-4，-5），·························································2分则解析式为20.yx···················································································3分（2）如图，求出点E的坐标为（-2，-10），点F的坐标（4，5）·······················4分分别过点E、F作EN⊥x轴于点N，FM⊥GM于点M，FM也垂直于x轴，证明△ENO≅△FMG，…………………………………………5分设点G的坐标为（m，n），则5-n=10，m-4=-2，则点G的坐标为（2，-5）；……………………………………6分（3）由于OE为定值，则只需求出OF的最小值即可，设点F的坐标为（a，20.a）根据勾股定理得，222222020()40.OFaaaa…7分显然当20.aa时，2OF最小，即a=25时，OF最小，OF=210，···················8分因此，当点F的坐标为（25，25）时，四边形OEGF周长最小，·················9分最小值为426410.·············································································10分24.（1）如图，易证△EBM1≅△EFN1，则∠EFN1=90°，则四边形BEFG为矩形，即FN1⊥AB；………………………………………3分（2）如图，同理，△EBM2≅△EFN2，则∠EFN2=90°，………5分由于∠EFN1+∠EFN2=180°，所以点N2在直线FN1上；……6分（3）易证四边形BEFG为正方形，易求BE=4；…………7分当点M1在线段AB的延长线上时，S1=1(42）xx=2122xx，此时x0；………………………………………………………9分当点M2在线段BA的延长线上时，①当3x4时，S2=211(4222xxxx）=……………………………………10分②当x4时，S3=211(4222）=xxxx，·················································11分