七年级数学参考答案及评分标准（阅卷前请认真校对，以防答案有误！）一、选择题（每小题3分，共24分）题号12345678答案C B C B B D C B二、填空题（每小题3分，共24分）9．1080°10．7011．5×510-12．613．1614．12015．416．140°或20°三、解答题（共72分）17．（本题满分6分）解：（1）原式＝1－4··························································2分＝－3．··································································································3分说明：02020＝1给1分，21()2-＝4给1分．（2）原式＝668a a -+···············································································5分＝67a -．·······························································································6分说明：23(2)a -＝68a -给1分，82a a ÷给1分．18．（本题满分6分）解：（1）原式＝21(6)3a ab -⨯-⋅⋅··········································2分＝32a b ．································································································3分说明：直接写出结果不扣分．（2）原式＝22362m mn mn n +--··································································5分＝22352m mn n +-．·················································································6分19．（本题满分6分）解：（1）原式＝224x -·······················································1分＝(4)(4)x x +-．·····················································································3分说明：直接写出结果不扣分．（2）原式＝2(21)ab a a -+··········································································4分＝2(1)ab a -．··························································································6分20．（本题满分6分）解：（1）如图，△A 1B 1C 1即为所求作．··································2分（2）平行································································································4分（3）如图，BP 即为所求作．·······································································6分说明：第（1）、（3）题中，不交待作图结果不扣分．21．（本题满分6分）解：原式＝2222444()5x xy y x y xy -+--+·····························2分＝2+5xy y ．·····························································································4分当x ＝6，y ＝－2时，原式＝－12＋20＝8．················································································6分CA B P 1C 1B 1A22．（本题满分6分）解：S ＝π×322－π×182·······················································3分＝π(322－182)＝700π（m 2）．························································································5分答：图中圆环的面积为700πm 2．································································6分23．（本题满分6分）解：（1）DE ∥AC ．····························································1分理由是：∵AD 平分∠BAC ，∴∠EAD ＝∠CAD ．·················································································2分∵∠EAD ＝∠EDA ，∴∠CAD ＝∠EDA ．∴DE ∥AC ．····························································································3分（2）∵在△ABC 中，∠BAC ＝95°，∠B ＝35°，∴∠C ＝180°－∠BAC －∠B ＝50°．······························································4分∵DE ∥AC ，∴∠EDF ＝∠C ＝50°．··············································································5分∵EF ⊥BC ，∴在△DEF 中，∠DEF ＝90°－∠EDF ＝40°．················································6分24．（本题满分8分）解：（1）2n (2n ＋2)＋1＝(2n ＋1)2．········································3分（2）∵左边＝2n (2n ＋2)＋1＝4n 2＋4n ＋1＝(2n ＋1)2＝右边，······························6分∴（1）中等式成立．（3）1000×1002＋1＝10012．·····································································8分25．（本题满分10分）解：（1）(a －b )2，(a ＋b )2－4ab ．········································2分说明：答对一个给1分，也可答成(a ＋b )2－4ab ，(a －b )2．（2）(a －b )2＝(a ＋b )2－4ab ；·······································································4分说明：写成(a ＋b )2－(a －b )2＝4ab 等其他变式也正确．（3）①1．·······························································································6分②解：由（2）得2()x y -＝2()4x y xy +-．∴2()x y -＝102－4×16＝36．····································································8分∴x －y ＝±6．························································································10分说明：答成x －y ＝6给1分．26．（本题满分12分）解：（1）∠1＋∠2＝2∠A ．···············································1分理由：连接AA'．∵∠1、∠2分别为△AEA'、△ADA'的外角，∴∠1＝∠EAA'＋∠EA'A ，∠2＝∠DAA'＋∠DA'A ．········································2分∴∠1＋∠2＝∠EAA'＋∠EA'A ＋∠DAA'＋∠DA'A ＝∠EAD ＋∠EA'D ．················3分由折叠得∠EAD ＝∠EA'D ．········································································4分∴∠1＋∠2＝2∠BAC ．·············································································5分说明：其他方法相应给分．（2）∠1－∠2＝2∠A ．··············································································6分理由：连接AA'，∵∠1、∠2分别△EAA ′、△DAA ′的外角，∴∠1＝∠EAA'＋∠EA'A ，∠2＝∠DAA'＋∠DA'A ．········································7分∴∠1－∠2＝∠EAA'＋∠EA'A －∠DAA'－∠DA'A ＝∠EAD ＋∠EA'D ．················8分由折叠得∠EAD ＝∠EA'D ．········································································9分∴∠1－∠2＝2∠A ．···············································································10分说明：其他方法相应给分．（3）55°．······························································································12分。