江苏省兴化市顾庄学区三校2019学年七年级下学期期末考试数学试卷【含答案及解析】39498.pdf

《江苏省兴化市顾庄学区三校2019学年七年级下学期期末考试数学试卷【含答案及解析】39498.pdf》由会员分享，可在线阅读，更多相关《江苏省兴化市顾庄学区三校2019学年七年级下学期期末考试数学试卷【含答案及解析】39498.pdf（20页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、2019 1.b?b3?b4A.b7 B.b7 C.-b8 D.b82.xy4kx-3y=-1kA.1 B.-1 C.2 D.-23.2x+1（）A.f l B.-5-i j*C.R 1 D.f*4.abA.-7 B.3C.D35.A.-2x-2,x 1 B.a2=b2,a3=b3 C.D.abbca/c6.ABPQBC ACPRABPSACRSAQ=PQ PR=PS,PA RPS AS=AR QP/AR BRPA CSP.A.4B.3 C.2 D.1 7.0.000004320.000004328.3x-4,2xx,5L9._ 10.ab,ac be _11.AB=DB,Z ABD=Z CB

2、E,14.120m4983xm,ymxy15.316.（a b）n（n 1234）nABCA DBE Z 2=90 ,AD=AE O1 1=2b 1 2 1 2a13 3 1(a+i)3-+3(+14 6 4 1 2+a “x-2 2017X2016171(1 2(1)2x2+x 5=0,x 12 x x 3+x+2x 218.12x3y 8xy2ii2x 1 X*1 19.1,3 2II I I II a_4-2 0 I 2 3 420.DABEACBECDO.1A=50,Z BOD=70C=30/B 2BOCA+Z B+ZC.21.1A3B263A2B29.AB 50AB3A22.ABCA

3、B=AC ADBCCEABAE=CE (AEFA CEB 2AF=2CD 23.xy m xv 0yv 0m24.mnO,ABOAm Bn.BAOABOPABAPB ABOAQ BQQ,APQBC,ABQ/C/Q/C 25.al:32-12=8x 1；a2:52-32=8x 2；a3:72-52=8X 3；_ a4 ann kakak+1 ak+2ABCk26.A=2 a-7 B=a2-4a+3 C=a2+6a-28 1B-A 0AB2BB=a2-4a+3=a2-4a+4-a-2 2-1=a-2+1 a-2-1=a-1 a-3 x2-4X-96 AC1CW：bxbb D.2H:J-1=3 f

4、JSkA5 3 4&(x+1)(x-3)=xax+b=itx2-2x-3,arg?b-36 D feW VRbatDi6-2x-2?7 2)S C1)APAR;TZ.ZlZAFQ!.ZFQS=Z 1+ZAPQ=2Z1 X/.FA ZEAC,/，ZBAC=2Z1J/.ZPQS=ZBACJ PQ#BRPZC3P,TPE=PS,ZFAR=ZP/SJ CPAS,AK=AS,_,_ ZMC=25AC,WfflHIii QBRPP HLC21 M1.FAT7TZBAC!(4)/PRF3AG 3)S-,I-iM iT 4.32XW ax 10-0000000324+32X4.32X 1 Ka10 08 93

5、,3xe,xx3 105 11UBDE=0BACBE=BCACB=0DEB.ABD=1CBE7 MBD DABE=3CBE+0ABE,iZIABCSDBEj AB=DBi fStaBDEEBAC,BE=BCP IaACE=UDE8,121 a+03ab=2,/.(a-b)ab-2alj=a+b+2ab4ab=(i+t)Mat,=0cl v0M,(2)J?=11 4 4 2E1-Aflr+x2-4=I2+K-3-5=0=5,3=2 218(1)2(K+2(a2)?125(z2)w(H 2 (L)4)(x+2)(x-2；(2)(y-4y+4)(IMI+4)=(y-2)(W 2)19-SS S&IF

6、&-W1.-9 Kiw -1 i 1 1 I I2(2z-1-9(5x+l)W&,4 2-15x36 15JCW-lhWll,11 20(1)30fl ZEOC=ZA+ZBZC,*SZBM!I 1/0OC,ZBEC(1)5ZC=30*,.ZHDO=eO j.00=70 30 ZEX=ZA+Z&+ZC.JBOC=ZBEC+ZC；ZBECZA+ZB,.ZBOC=ZA+ZB+ZC*211 5X I 1 BFBI572)5ME50)m3(50-ra),m37.5.A122 CE-LABISAFtBiEFEBj AF.BC,BECl）ADxBC/CE-LAB iCFD=iB MFDSFE*AFE=xB

7、ftEER 3AFE=HB）QAEF=0CEBAE=CE iAEF-iCEB（AAS）J 2）AB=AC ADiBC BG=2Cm iAEF-iCEB AF=BG AF=2CD-23（1）AD_BC (L)X25x=iOm-5,Em-172+y=5m+6,Jc-2y=-17x 0X2+05K=10m-5,K=2m-lj 212n-l-217,y=mC，inK-Sj nJLVT-24 (1)/AFiJSZMOZABOZAOB90e J.ZAFB=L8O0:;MB1 1(ZOAB-hAEOi=180a-X9(T=135a*2)TBQB0BP4.ZCA(J=ZQBP=9O*,yZAPB1354,/-

8、ZQ=45Q，/.ZC=45e 25(1)97=8X4j (2n+l)(2n-l)C3)41b 2)(2nM)2-(2n-l)Mn (Sn+l)-(2n-l)s=E(2n+l)+(2n-l)(2n41)-(2n.-l：=4nX2=9nj(2)a.-kj(k+L),(lc+2)8 bBffiQi 268 oiMMg S a?si (1)BA;(2)(X48)(x-12)3 2aC；#3A=C;a3,A0,BA;(2)(Dx2-4x-96=x2-4x+4-100=(x-2)2-10=(x-2+10)(x-2-10)=(x-t8)(x-12);C-A=a+6a282a47=aS4a21=(a+-7)(a-3).a2,a470,2aC；3A=C;a3AC.