大兴区2023～2024学年度第二学期初三期末检测数学参考答案及评分标准一、选择题（共16分，每题2分）题号12345678答案DBDBCBCA二、填空题（共16分，每题2分）题910111213141516号答()()答案不唯一，x53x+1x−1x=−1>6238，4案例如：AB=CD三、解答题（共68分，第17-20题，每题5分，第21题6分，第22-23题，每题5分，第24-26题，每题6分，第27～28题，每题7分）解答应写出文字说明、演算步骤或证明过程.317．解：原式23232·························································4分231.············································································5分2(x−1)x+3,①18.解：4x+1x.②2解不等式①，得x<5.································································2分1解不等式②，得x＞−.·····························································4分21所以原不等式组的解集为−x5.·············································5分2x2y22xy2x19.解：原式.xxxyx2y22xy2x.……………………………………….1分xxy（xy）22x.……………………………..…………………2分xxy2（xy）.………………………………………………3分∵xy50.…………………………………………………………………4分∴xy5.∴原式=10......………………………….………………………………...…5分初三数学参考答案及评分标准（）20.方案一：解：由题意得，DEFABC90，DF∥AC.…………………1分DFEACB.…………………2分DEF∽ABC.…………………3分DEEF.…………………4分ABBCDE1.8,EF0.9,BC6，1.80.9.AB6AB12.答：旗杆高度为12m.……………………………………5分方案二：解：由题意得，DECABC90，DCEACB………1分DEC∽ABC.……………………………2分DEEC.……………………………3分ABBCDE1.6,EC1.2,BC9，1.61.2.………………………………4分AB9AB12.答：旗杆高度为12m.……………………………5分21.（1）证明：四边形ABCD是平行四边形，AD=BC，AD∥BC.点E,F分别为AD,BC中点，11AF=AD，EC=BC.22AF=EC.四边形AECF是平行四边形.………………………………………………….1分∠BAC=90°，点E为BC中点，1AE=BC=EC.2四边形AECF是菱形.….………………………………………………….2分初三数学参考答案及评分标准（）（2）解：连接EF，交AC于点O.在Rt△ABC中，AB2+AC2=BC2，AB=6,EC=10,AC=8（舍负）.….…………………………………………….3分1AE=BC,2AE=5.AE=AG，AG=5.….……………………………..………………….4分四边形AECF是菱形，O是AC的中点，AC⊥EF.11AO=AC=4，EO=AB=3.22OG=AG−AO=5−4=1.….………………………………………………….5分在Rt△EOG中，EO2+OG2=EG2，EG=10（舍负）.….…………………………………………..…….6分22.解：（1）m170，n169.5；……………………………………………………2分（2）B队；……………………………………………………3分（3）①.……………………………………………………5分23.解：（1）∵函数y=2xy=kx+b的图象是由的图象平移得到的，∴k=2.……………………………………………...………………………………………1分把（1，5）代入y=2x+b，解得b=3.…………………………………………………2分∴函数的表达式是y=2x+3.…………………