廣西柳州市2019年中考數(shù)學(xué) 專(zhuān)題訓(xùn)練02 多結(jié)論題.doc

專(zhuān)題訓(xùn)練(二)多結(jié)論題1.xx遵義如圖ZT2-1,拋物線(xiàn)y=ax2+bx+c經(jīng)過(guò)點(diǎn)(-1,0),對(duì)稱(chēng)軸l如圖所示.則下列結(jié)論:abc>0;a-b+c=0;2a+c<0;a+b<0,其中所有正確的結(jié)論是()圖ZT2-1A.B.C.D.2.如圖ZT2-2,拋物線(xiàn)y=ax2+bx+c(a0)的對(duì)稱(chēng)軸為直線(xiàn)x=-2,與x軸的一個(gè)交點(diǎn)在(-3,0)和(-4,0)之間,其部分圖象如圖所示,則下列結(jié)論:4a-b=0;c<0;-3a+c>0;4a-2b>at2+bt(t為實(shí)數(shù));點(diǎn)-92,y1,-52,y2,-12,y3是該拋物線(xiàn)上的點(diǎn),則y1<y2<y3.正確的有()圖ZT2-2A.4個(gè)B.3個(gè)C.2個(gè)D.1個(gè)3.如圖ZT2-3,在矩形ABCD中,點(diǎn)E,F分別在邊AB,BC上,且AE=13AB,將矩形沿直線(xiàn)EF折疊,點(diǎn)B恰好落在AD邊上的點(diǎn)P處,連接BP交EF于點(diǎn)Q,對(duì)于下列結(jié)論:EF=2BE;PF=2PE;FQ=4EQ;PBF是等邊三角形.其中正確的是()圖ZT2-3A.B.C.D.4.如圖ZT2-4,在正方形ABCD中,BPC是等邊三角形,BP,CP的延長(zhǎng)線(xiàn)分別交AD于點(diǎn)E,F,連接BD,DP,BD與CF相交于點(diǎn)H,給出下列結(jié)論:BE=2AE;DFPBPH;PFDPDB;DP2=PHPC.其中正確的是()圖ZT2-4A.B.C.D.5.xx宜賓如圖ZT2-5,在矩形ABCD中,AB=3,CB=2,點(diǎn)E為線(xiàn)段AB上的動(dòng)點(diǎn),將CBE沿CE折疊,使點(diǎn)B落在矩形內(nèi)點(diǎn)F處,下列結(jié)論正確的是.(寫(xiě)出所有正確結(jié)論的序號(hào))圖ZT2-5當(dāng)E為線(xiàn)段AB中點(diǎn)時(shí),AFCE;當(dāng)E為線(xiàn)段AB中點(diǎn)時(shí),AF=95;當(dāng)A,F,C三點(diǎn)共線(xiàn)時(shí),AE=13-2133;當(dāng)A,F,C三點(diǎn)共線(xiàn)時(shí),CEFAEF.6.xx南充如圖ZT2-6,正方形ABCD和正方形CEFG的邊長(zhǎng)分別為a和b,正方形CEFG繞點(diǎn)C旋轉(zhuǎn).給出下列結(jié)論:BE=DG;BEDG;DE2+BG2=2a2+2b2.其中正確的結(jié)論是(填寫(xiě)序號(hào)).圖ZT2-6參考答案1.D解析 開(kāi)口向下,a<0.對(duì)稱(chēng)軸與x軸的正半軸相交,a,b異號(hào),即b>0.拋物線(xiàn)與y軸正半軸相交,c>0,即abc<0,結(jié)論錯(cuò)誤.拋物線(xiàn)y=ax2+bx+c經(jīng)過(guò)點(diǎn)(-1,0),a-b+c=0,結(jié)論正確.當(dāng)x=2時(shí),y<0,即4a+2b+c<0,又b=a+c,4a+2(a+c)+c<0,即2a+c<0,結(jié)論正確.c=b-a,a+b<0,結(jié)論正確.2.C解析 拋物線(xiàn)y=ax2+bx+c(a0)的對(duì)稱(chēng)軸為直線(xiàn)x=-2,-b2a=-2,4a-b=0,故正確;拋物線(xiàn)y=ax2+bx+c(a0)的對(duì)稱(chēng)軸為直線(xiàn)x=-2,與x軸的一個(gè)交點(diǎn)在(-3,0)和(-4,0)之間,另一個(gè)交點(diǎn)位于(-1,0)和(0,0)之間,拋物線(xiàn)與y軸的交點(diǎn)在原點(diǎn)的下方,c<0.故正確;4a-b=0,b=4a.拋物線(xiàn)y=ax2+bx+c(a0)與x軸有兩個(gè)交點(diǎn),=b2-4ac=(4a)2-4ac=16a2-4ac>0.a<0,4a-c<0,c>4a,-3a+c>-3a+4a=a<0,故錯(cuò)誤;4a-b=0,b=4a,at2+bt-(4a-2b)=at2+4at-(4a-24a)=at2+4at+4a=a(t2+4t+4)=a(t+2)2.t為實(shí)數(shù),a<0,a(t+2)20,at2+bt-(4a-2b)0,at2+bt4a-2b,即4a-2bat2+bt,錯(cuò)誤;點(diǎn)-92,y1,-52,y2,-12,y3是該拋物線(xiàn)上的點(diǎn),將它們描在圖象上如圖:由圖象可知:y1<y3<y2,錯(cuò)誤.綜上所述,正確的有2個(gè).故選C.3.D解析 AE=13AB,AB=3AE,BE=2AE.由翻折的性質(zhì)得,PE=BE.APE=30,AEP=90-30=60,BEF=12(180-AEP)=12(180-60)=60,EFB=90-60=30,EF=2BE,故正確;BE=PE,EF=2PE.EF>PF,PF<2PE,故錯(cuò)誤;由翻折可知EFPB,EBQ=EFB=30,BE=2EQ,EF=2BE,FQ=3EQ,故錯(cuò)誤;由翻折的性質(zhì)知,EFB=EFP=30,BFP=30+30=60.PBF=90-EBQ=90-30=60,PBF=PFB=60,PBF是等邊三角形,故正確.綜上所述,結(jié)論正確的是.4.C解析 在正方形ABCD中,A=90.由BPC是等邊三角形,可得CBP=60,ABP=30,BE=2AE,即正確;BD是正方形ABCD的對(duì)角線(xiàn),可得BCD是等腰直角三角形,CBD=CDB=45,可得PBD=15.CD=CP=CB,PCD=30,可得CPD=CDP=75,BPD=75+60=135,FDP=90-75=15,PFD=90-PCD=90-30=60,FPD=180-PDF-PFD=180-15-60=105,PBD=PDF,BPH=DFP,DFPBPH,即正確;BPDDPF,PFDPDB錯(cuò)誤;由PDH=PDC-CDB=75-45=30=PCD,CPD=DPH,可得PDCPHD,DP2=PHPC,即正確.5.解析 由折疊的性質(zhì)可知CF=CB,CFE=90,CEB=CEF,當(dāng)E為AB中點(diǎn)時(shí),BE=EF=AE=32,FAE=AFE,FEB=FAE+AFE,CEB=CEF=FAE=AFE,AFCE,故正確;E為AB中點(diǎn)時(shí),BE=32,BC=2,CE=52,過(guò)點(diǎn)E作EMAF于點(diǎn)M,AFE=FEC,EMAF,CFE=90,AF=2MF,MFEFEC,MFEF=EFEC,即MF32=3252,MF=910,AF=95,故正確;當(dāng)A,F,C三點(diǎn)共線(xiàn)時(shí),AFE=90,AC=22+32=13,設(shè)BE=x,則EF=x,AE=3-x,AF=13-2,在RtAFE中,(13-2)2+x2=(3-x)2,解得x=213-43,AE=3-x=13-2133,故正確;AF=13-2,CF=2,AFCF,錯(cuò)誤.6.解析 正方形的各邊相等,各角都是90,CB=CD,CE=CG,BCD=ECG=90.BCD+DCE=ECG+DCE,即BCE=DCG.BCEDCG(SAS),BE=DG.結(jié)論正確.如圖,設(shè)BE交DC于點(diǎn)M,交DG于點(diǎn)O.由BCEDCG可知CBE=CDG.又BMC=DMO,DOB=DCB=90,即BEDG.結(jié)論正確.連接BD,EG.BEDG,DE2+BG2=(OD2+OE2)+(OB2+OG2)=(OD2+OB2)+(OE2+OG2)=BD2+EG2.由勾股定理得BD2+EG2=2a2+2b2.DE2+BG2=2a2+2b2.結(jié)論正確.綜上所述,正確的結(jié)論是.