学法大视野·数学·九年级上册(湘教版)·答案

-课时参考答案kxBkkx0)的形式.22xxkkx2x133-3解得{EQ\*jc3\*hps31\o\al(\s\up12(k),k)EQ\*jc3\*hps22\o\al(\s\up4(1),2)22xxEQ\*jc3\*hps24\o\al(\s\up8(0),8)EQ\*jc3\*hps24\o\al(\s\up8(0),5)122-xxxEQ\*jc3\*hps31\o\al(\s\up9(y),y)xEQ\*jc3\*hps24\o\al(\s\up8(k),1)xEQ\*jc3\*hps31\o\al(\s\up13(y),y)解得{EQ\*jc3\*hps31\o\al(\s\up12(x),y)EQ\*jc3\*hps26\o\al(\s\up4(1),1)或{.,-55kkk3xxEQ\*jc3\*hps24\o\al(\s\up9(k),x)2x2xEQ\*jc3\*hps24\o\al(\s\up9(m),x)-xxx2-aaa,aEQ\*jc3\*hps24\o\al(\s\up8(1),2)EQ\*jc3\*hps24\o\al(\s\up8(1),2)EQ\*jc3\*hps24\o\al(\s\up8(4),a)m221.理由如下:-EQ\*jc3\*hps24\o\al(\s\up8(5),2)1.EQ\*jc3\*hps24\o\al(\s\up7(P),F)PF,F22x22xEQ\*jc3\*hps22\o\al(\s\up4(1),2)EQ\*jc3\*hps31\o\al(\s\up11(+),k1)EQ\*jc3\*hps31\o\al(\s\up10(k),b).EQ\*jc3\*hps24\o\al(\s\up7(1),2)-22122EQ\*jc3\*hps24\o\al(\s\up9(1),2)kkx4x8x8xEQ\*jc3\*hps31\o\al(\s\up9(k),b)EQ\*jc3\*hps31\o\al(\s\up9(1),2)8EQ\*jc3\*hps31\o\al(\s\up18(y),y)8xEQ\*jc3\*hps31\o\al(\s\up11(x),y)EQ\*jc3\*hps31\o\al(\s\up11(x),y)x88x88x88-EQ\*jc3\*hps31\o\al(\s\up10(a),b)EQ\*jc3\*hps24\o\al(\s\up8(1),2)22x2m222-EQ\*jc3\*hps31\o\al(\s\up14(m),m)22所以此方程不可能为一元二次方程.-2,92,2,EQ\*jc3\*hps24\o\al(\s\up9(3),2)EQ\*jc3\*hps24\o\al(\s\up9(1),2)2.EQ\*jc3\*hps24\o\al(\s\up7(1),2)EQ\*jc3\*hps24\o\al(\s\up7(1),4)EQ\*jc3\*hps24\o\al(\s\up7(9),16)EQ\*jc3\*hps24\o\al(\s\up7(1),4)EQ\*jc3\*hps24\o\al(\s\up7(9),16)EQ\*jc3\*hps24\o\al(\s\up8(1),4)EQ\*jc3\*hps24\o\al(\s\up8(3),4)EQ\*jc3\*hps24\o\al(\s\up8(1),4)EQ\*jc3\*hps24\o\al(\s\up8(3),4)12EQ\*jc3\*hps24\o\al(\s\up8(1),2)222222-2225454545432=+2224即2即EQ\*jc3\*hps24\o\al(\s\up9(5),4)EQ\*jc3\*hps24\o\al(\s\up9(7),4)EQ\*jc3\*hps24\o\al(\s\up9(1),2)222p2p2-2即p2-4q22即4,EQ\*jc3\*hps24\o\al(\s\up3(p),2)EQ\*jc3\*hps24\o\al(\s\up6(3),2)EQ\*jc3\*hps24\o\al(\s\up6(3),4)EQ\*jc3\*hps24\o\al(\s\up6(1),2)EQ\*jc3\*hps24\o\al(\s\up6(3),4))2,EQ\*jc3\*hps24\o\al(\s\up7(3),4)EQ\*jc3\*hps24\o\al(\s\up7(1),16)EQ\*jc3\*hps24\o\al(\s\up9(1),2)16-6166-1EQ\*jc3\*hps24\o\al(\s\up9(1),12)EQ\*jc3\*hps24\o\al(\s\up9(17),12)1EQ\*jc3\*hps24\o\al(\s\up9(3),2)2EQ\*jc3\*hps24\o\al(\s\up9(4),3)4412=+2242499---2EQ\*jc3\*hps24\o\al(\s\up9(1),2)22222-1 1EQ\*jc3\*hps22\o\al(\s\up4(√),2)EQ\*jc3\*hps22\o\al(\s\up4(√),2)EQ\*jc3\*hps22\o\al(\s\up6(√),2)22EQ\*jc3\*hps28\o\al(\s\up8(5),6)EQ\*jc3\*hps28\o\al(\s\up8(5),6)EQ\*jc3\*hps28\o\al(\s\up8(7),6)2EQ\*jc3\*hps24\o\al(\s\up9(1),3)-2EQ\*jc3\*hps22\o\al(\s\up4(+√),4)EQ\*jc3\*hps22\o\al(\s\up4(√),4)EQ\*jc3\*hps24\o\al(\s\up5(+),2)EQ\*jc3\*hps22\o\al(\s\up4(-√),2)-所以方程有两个相等的实数根,EQ\*jc3\*hps24\o\al(\s\up2(2),2)==.EQ\*jc3\*hps24\o\al(\s\up9(3),2)12EQ\*jc3\*hps24\o\al(\s\up5(2),2)EQ\*jc3\*hps24\o\al(\s\up5(2±2),2)√2,2EQ\*jc3\*hps24\o\al(\s\up6(±),2)EQ\*jc3\*hps24\o\al(\s\up6(20),1)2-EQ\*jc3\*hps22\o\al(\s\up4(√),2)222EQ\*jc3\*hps24\o\al(\s\up8(13),3)EQ\*jc3\*hps24\o\al(\s\up9(4),3)2EQ\*jc3\*hps24\o\al(\s\up9(7),3)2222-22222(3)用因式分解法:方程两边分别分解因22EQ\*jc3\*hps22\o\al(\s\up4(√),6)22EQ\*jc3\*hps24\o\al(\s\up9(8),5)2-25-2+√2-2-√222-222222121252525252-aaab-28解:`方程有两个不相等的实数根,1EQ\*jc3\*hps24\o\al(\s\up9(5),2)EQ\*jc3\*hps24\o\al(\s\up8(5),2)2是方程的两个实数根,EQ\*jc3\*hps24\o\al(\s\up8(3),2)EQ\*jc3\*hps24\o\al(\s\up8(-m),2)x1x2x1x2:m的值为2.-EQ\*jc3\*hps24\o\al(\s\up9(m),m)EQ\*jc3\*hps24\o\al(\s\up9(1),1)22答:捐款增长率为10%.EQ\*jc3\*hps24\o\al(\s\up8(4),0)EQ\*jc3\*hps24\o\al(\s\up8(x),1)-EQ\*jc3\*hps24\o\al(\s\up9(x),2)2222212212222EQ\*jc3\*hps24\o\al(\s\up8(1),2)2经检验,它们均符合题意.2-12答:矩形花园各角处的正方形观光休息亭的边长为4422EQ\*jc3\*hps24\o\al(\s\up9(1),2)2222-2EQ\*jc3\*hps24\o\al(\s\up9(2),3)3y3yy3x2x2y3x7.y4ABACBC3,AB+AC+BC3EQ\*jc3\*hps24\o\al(\s\up7(x),3)EQ\*jc3\*hps24\o\al(\s\up7(y),5)EQ\*jc3\*hps24\o\al(\s\up7(z),7)-x+y-z3k+5k-7kk.x+y-z3k+5k-7kk.33n3n31.C2.C3.D4.C5.AEQ\*jc3\*hps24\o\al(\s\up7(b),b)EQ\*jc3\*hps24\o\al(\s\up7(4k),2k)EQ\*jc3\*hps24\o\al(\s\up7(12),4k)EQ\*jc3\*hps24\o\al(\s\up7(17k),3k)EQ\*jc3\*hps24\o\al(\s\up7(17),3)EQ\*jc3\*hps24\o\al(\s\up11(a),b)EQ\*jc3\*hps24\o\al(\s\up11(C),d)EQ\*jc3\*hps24\o\al(\s\up11(e),f)EQ\*jc3\*hps24\o\al(\s\up11(2),3)2b-d-5f3.2b-d-5f3.2b-d-5f32b-d-5f3CDnCDnaCEQ\*jc3\*hps24\o\al(\s\up7(BC),AC)12-x4ABACx6-DBECEQ\*jc3\*hps24\o\al(\s\up8(d),b)EQ\*jc3\*hps24\o\al(\s\up8(a),C)ABACX+33EQ\*jc3\*hps24\o\al(\s\up9(AC),AB)EQ\*jc3\*hps24\o\al(\s\up9(CB),AC)EQ\*jc3\*hps24\o\al(\s\up9(X),3)AB2AB2—EQ\*jc3\*hps24\o\al(\s\up8(5),2)EQ\*jc3\*hps24\o\al(\s\up7(BC),AB)EQ\*jc3\*hps24\o\al(\s\up7(5),2)—EQ\*jc3\*hps24\o\al(\s\up8(5),2)-1.B2.D3.C4.B5.B√5-13-√5√5-13-√5,在直角三角形BCE中,,AB√29BC√26AB√29BC√26bbEQ\*jc3\*hps31\o\al(\s\up0(=),k)EQ\*jc3\*hps31\o\al(\s\up19(k),7)=4k,22,,所以△ABC是直角三角形.5DF,ACDF,ACDF,AB3AB3BC2ACAB3AB3BC2AC5DF5,DF5,-3552ACX-3XDEQ\*jc3\*hps24\o\al(\s\up9(AD),AB)EQ\*jc3\*hps24\o\al(\s\up9(AE),AC)EQ\*jc3\*hps24\o\al(\s\up9(3),5)3EQ\*jc3\*hps24\o\al(\s\up9(AD),BD)EQ\*jc3\*hps24\o\al(\s\up9(CF),BF)EQ\*jc3\*hps24\o\al(\s\up9(4),8)ABACABACADACADACABADABAD-EFGHFG42ABCDBC189,5.解:因为梯形AEFD∽梯形EBCF,ADEFAEEFBCEB,EFBCEB,AEAD42EBEF63EBEF632-:EFEHEFEHFGEHFGEHAPPQ解::四边形ABCD是平行四边形,EQ\*jc3\*hps24\o\al(\s\up9(BE),CD)EQ\*jc3\*hps24\o\al(\s\up9(1),3)EQ\*jc3\*hps24\o\al(\s\up9(BE),AE)EQ\*jc3\*hps24\o\al(\s\up9(1),4)EQ\*jc3\*hps24\o\al(\s\up8(CD),AE)EQ\*jc3\*hps24\o\al(\s\up8(3),4)解:△ABC△ADE,理由如下:::又:在△AOB与△COD中,:-EQ\*jc3\*hps24\o\al(\s\up8(BD),AB)EQ\*jc3\*hps24\o\al(\s\up8(CD),BC)EQ\*jc3\*hps24\o\al(\s\up8(AD),AC)EQ\*jc3\*hps24\o\al(\s\up8(CD),AD)2—EQ\*jc3\*hps24\o\al(\s\up7(5),2)-(2)四边形AFCE是菱形,理由如下::即AC、EF互相垂直平分,:四边形AFCE是菱形.ABBCABAC变式训练1-2:证明::四边形ABCD是正方形,:BP〓3PC,:CM:PC〓2:1,ADCM:△ADM△MCP.解:相似.理由如下:-DE2DF2EF2DE2DF2EF2ABACBC∴DE、DF、EF分别为△ABC的中位线,CBCABA2CBCABA2DFEFDE2DFEFDE2BCABBCABACOAACOAB'C'A'B'ACBCABACBCAB-MCBCADAE,AD·AC20×9AB15AB15SAB29EQ\*jc3\*hps19\o\al(\s\up5(团ABC),团ADE)SAB29DE1-EQ\*jc3\*hps24\o\al(\s\up8(DM),BM)EQ\*jc3\*hps24\o\al(\s\up8(DE),BF)3S△EDM=S△FBMDEBFDFFGBFBGABBFABBGDB+3BD+7AB12AB12DC-EQ\*jc3\*hps24\o\al(\s\up7(AB),DE)EQ\*jc3\*hps24\o\al(\s\up7(BC),CD)BCDE,BCDE,7090,88PCCB,PCCB,答:点光源S与平面镜的距离SA的长是12cEFDE,EFDE,.A'B'62-AB1AB4,AB4,解:(1)如图所示.5.解:(1)OAE与△OBF相似.理由:EQ\*jc3\*hps24\o\al(\s\up8(OA),OB)EQ\*jc3\*hps24\o\al(\s\up8(OC),OD)EQ\*jc3\*hps24\o\al(\s\up8(OE),OF)EQ\*jc3\*hps24\o\al(\s\up8(OC),OD)OBOFOBOF△OAE与△OBF位似.理由:-,又△OAE和△OBF对应点的连线都经过点O,(2)△ACE与△BDF位似.理由:EQ\*jc3\*hps24\o\al(\s\up8(AE),BF)EQ\*jc3\*hps24\o\al(\s\up8(OA),OB)EQ\*jc3\*hps24\o\al(\s\up8(AC),BD)EQ\*jc3\*hps24\o\al(\s\up8(OA),OB)EQ\*jc3\*hps24\o\al(\s\up8(OC),OD)EQ\*jc3\*hps24\o\al(\s\up8(CE),DF)EQ\*jc3\*hps24\o\al(\s\up8(OC),OD)BDDFBF又△ACE和△BDF对应点的连线都经过点O,∴△ACE与△BDF位似.333或或222为位似中心,为位似中心,EQ\*jc3\*hps24\o\al(\s\up10(AB),AB)EQ\*jc3\*hps24\o\al(\s\up10(AD),AD)ABADEQ\*jc3\*hps24\o\al(\s\up8(AB),AD)EQ\*jc3\*hps24\o\al(\s\up8(2),1)-2,EQ\*jc3\*hps24\o\al(\s\up9(BC),AB)EQ\*jc3\*hps24\o\al(\s\up9(6),10)EQ\*jc3\*hps24\o\al(\s\up9(3),5)EQ\*jc3\*hps24\o\al(\s\up9(AC),AB)EQ\*jc3\*hps24\o\al(\s\up9(8),10)EQ\*jc3\*hps24\o\al(\s\up9(4),5)532—245EQ\*jc3\*hps24\o\al(\s\up9(1),2)1.C2.B3.C4.C5.B6.0.6421-EQ\*jc3\*hps24\o\al(\s\up9(1),2)EQ\*jc3\*hps24\o\al(\s\up8(D),B)EQ\*jc3\*hps24\o\al(\s\up8(21),4)AD√21AD√21CABACACBCABABABEQ\*jc3\*hps24\o\al(\s\up8(AC),AB)EQ\*jc3\*hps24\o\al(\s\up8(15x),17x)EQ\*jc3\*hps24\o\al(\s\up8(15),17)AC15BC8-D—24+2+22√6√6—2,5-EQ\*jc3\*hps24\o\al(\s\up8(BC),OB)EQ\*jc3\*hps24\o\al(\s\up8(3),5)EQ\*jc3\*hps24\o\al(\s\up8(C),B)EQ\*jc3\*hps22\o\al(\s\up8(√),5)bbABB一-4√33√33CD3EQ\*jc3\*hps24\o\al(\s\up9(AC),BC)EQ\*jc3\*hps24\o\al(\s