鄞州区初中数学中考模拟考试参考答案和评分标准注：17题有错误答案出现扣4分，漏下一个正确答案扣2分。

三、解答题：（第19题6分，第20、21题8分，第22～24题各10分，第25题12分，第26题14分，共78分）19． 解：原式=)12(322++-+x x x x ＝)12(322++-+x x x x ＝1-x ··············· 4分 当1x =时，原式=2 ································································· 6分20．（1） ······························································· 4分（2） ····························································· 8分21.（1）略（人数为14人） ············································································ 2分（2）126---------------------------------------------------------------------------------------- ----4分（3）两名学生选取A ，B ，C ，D 的树状图如图所示∴选中同一名著的概率为P ＝41164=································································ 8分 A BB B B CC D甲乙DD A DC A A C A DC B22．解： （1）将B 的坐标（－3，－2）代入y 2＝kx得k =6,∴y 2＝6x ························ 2分∴A 的坐标为（1,6） ···················································· 3分将A （1,6），B （－3，－2）分别代入y 1＝ax ＋b 得a =2，b =4 ∴y 1＝2x ＋4 ···················································································· 5分 （2）令y 1＝0，则2x ＋4=0，∴x =-2，∴C （－2，0） ··································· 7分∴当601<<y 时x 的取值范围解12<<x - ····································· 10分23．解：（1）连结BD 交AC 于点O ，∵四边形ABCD 为正方形，∴OA =OB =OC =OD ,AC ⊥BD ， 又∵AE =CF , ∴OE ＝OF ，∴四边形BEDF 为平行四边形， ························································ 3分 ∵EF 垂直平分BD ，∴EB =ED , ∴四边形BEDF 是菱形. ············································································· 5分 （2）过点E 作EH ⊥AB ，垂足为H ，设EH =x ， ∵正方形边长为4，∴AC =BD =24∵∠BAE =45〫，31tan =∠ABE∴AH =x ,BH = 3x ,∴x +3x = 4，∴x =1， ························································ 7分 ∴AE =CF =2,∴EF =22∴菱形BEDF 的面积为82222421=⨯=⋅EF BD ····································· 10分 24．解：（1）设该网店3M 1860口罩每袋的售价为x 元，3M 8210口罩每袋的售价为y 元，根据题意得：523110x y x y -=⎧⎨+=⎩，解这个方程组得：2520x y =⎧⎨=⎩，故该网店3M 1860口罩每袋的售价为25元，3M 8210口罩每袋的售价为20元 ············· 3分（2）设该网店购进3M 1860口罩m 袋，则购进3M 1860口罩(500－m )袋，根据题意得：4(500)522.418(500)10000m m m m ⎧>-⎪⎨⎪+-≤⎩， 解这个不等式组得：22292＜m ≤ 227113，因m 是整数，故有5种进货方案. ·············· 7分 （3）设网店获利为w 元，则有w ＝(25－22.4)m ＋(20－18)(500－m )＝0.6m ＋1000，因w 随m 的增大而增大，故当m ＝227时，w 最大， W 最大＝0.6×227＋1000＝1136.2（元）.故网店购进3M 1860口罩227袋，3M 1860口罩273袋时，获利最大为1136.2元. ······· 10分25．解：（1）∵直线3-=kx y 与y 轴交于点B ，∴B （0，-3）∴3)10(2-=+-m∴4-=m ··················································································· 2分由 得A (3，0),C (-1，0) ∴3k -3= 0,∴k =1. ······································································· 4分 （2）①∵BD ∥x 轴且B ，D 都在抛物线 上， ∴点D 与点B 关于对称轴直线x =1对称，即D 为（2，-3） ∴直线CD 的解析式为y =-x -1,∴点E 为（0，-1） ····························· 6分 ∴ΔOCE 为等腰直角三角形∴∠BCD +∠OBC =∠CEO =45〫 ························································ 8分 ②当直线BP 与x 轴的交点F 在点A 的左侧时， ∵A (3，0),B （0，-3），∴ΔAOB 为等腰直角三角形， ∴∠ABF +∠OBF =∠ABO =45〫∵∠BCD +∠OBC =45〫且∠ABP =∠BCD ∴∠OBF =∠OBC ∴ΔOBF ≌ΔOBC ∴OF =OC =1，即F （1,0） ∴直线BF 的解析式为y =3x -3 联立方程组得{33322-=--=x y x x y ，解得{1203211==-=y x y ∴P 点坐标为（5,12） ································································· 10分当直线BP 与x 轴的交点F 在点A 的右侧时， ∵∠ABF +∠DBF =∠ABD =45〫 ∵∠BCD +∠OBC =45〫且∠ABF =∠BCD ∴∠DBF =∠OBC ∵∠DBF =∠OFB ∴∠OFB =∠OBC ∴tan ∠OFB =tan ∠OBC =31 ∴OF =9，即F （9,0） ∴直线BF 的解析式为y =31x -3 联立方程组得⎪⎩⎪⎨⎧-=--=331322x y x x y ，解得{⎪⎩⎪⎨⎧===-=3792032211(x y x y -舍）， ∴P 点坐标为)920,37(-··································································· 12分 04)1(2=--x 4)1(2--=x y26．解：（1）正方形 ······················································································· 3分（2）连结BD ，AE ，∵∠BAC =90°∴BD 为⊙O 的直径，∴∠BED =∠CED =90〫，∵四边形ABED 为圆美四边形,∴BD ⊥AE ， ∴∠ABD +∠BAE =90〫， ∵∠CAE +∠BAE =90〫∴∠ABD =∠CAE ，∴弧AD =弧DE ，∴AD =DE ∴在等腰直角ΔCDE 中DE CD 2=,∴AD CD 2=,∴AD AC )12(+=,∴12+=DEAB··········································（3）①∵P A ⊥PD ，PB ⊥PE ,∴∠APD =∠BPE =90〫,∵∠PBC =∠ADP ,∴ΔAPD ∽ΔEPB , ··················································· 8分 ∴PB PD EP AP =,∴PBEP PD AP =, 又∵∠APD +∠DP E=∠BPE +∠DPE ,即∠APE =∠DPB∴ΔAPE ∽ΔDPB , ······························∴∠AEP =∠DBP ,又∵∠DBP +∠PGB =90〫,∠PGB =∠EGF ， ∴∠AEP +∠EGF =90〫即∠BFE =90〫，∴BD ⊥AE ，又∵A ，B ，E ，D ∴四边形ABED 为圆美四边形 ··················②∵BD ⊥AE ，∴AD 2+BE 2=AF 2+FD 2+BF 2+EF 2，AB 2+DE 2=AF 2+BF 2+DF 2+EF 2 ∴AD 2+BE 2=AB 2+DE 2， ··································································· 12分 ∵A ，B ，E ，D 在同一个圆上，∴∠CDE =∠CBA ,∵∠C =∠C ,∴ΔCDE ∽ΔCBA , ∴33==BC CD AB DE ········································································ 13分 设P A =x ,PE =8-x ，DE =y ，AB =y 3∵α=30〫,∠APD =∠BPE =90〫∴AD =2x ，BE =2(8-x )， ∴2222)]8(2[)2()3(x x y y -+=+，∴32)4(264162)8(22222+-=+-=-+=x x x x x y ，∵2>0∴当x =4时y 取到最小值24即DE 的最小值为24 ······························ 14分。