1、（第2題）6 .已知一個等腰三角形兩內(nèi)角的度數(shù)之比為1 : 4 ,則這個等腰三角形頂角的度數(shù) 為.7 .如圖，AC、BD相交于點O, / A=/D,請你再補(bǔ)充一個條件，使得AOBADOC,你補(bǔ)充的條件是.（第7題）oU 'ceated, and care masses if, f_swok, i n del hivestgaton, moreout boui _e maserpiee, ms ofesearce ls more t o it e d of de cS on vs i n, more to in natonai some has fec ofnewspaper S han

2、g pu.hed,“ adva wk, a nd pUUy cag pay due Thid,ifmaionaubmi- to be pragma, c Quck a nd tmei y The a-nts sd： fr the tme sysem； saemetback for te time be ngg Therefore, the sums- of ifrm.ntdfu", tat s, Ind theprob-a . edti r_ wiigsed a. rea. faser approval ai d eeda ck timpiemet quck y T be t ue

3、and uae Tue many elec the ul picue of - nt* one s one w o, ti s s the ie of the infmain. Accuicy is prmar. a nd .a-e . - stto reort, faSe cam and .mmig Iveste 2. sue-on ad isst on, aound ad protecig the i ite s t tuch the tut, pracicl esuts A dhee t pepe ore "d t he most mpor.nt ting s t e , s

4、.ad aI d devl ip te fudme nal iterss of he o'er «i ng mjri y of te pepe. cay out ispecin, so mus go de e among te m-ss, god_ pitte rrali a , s pay atetinttepeopC iood, tgas t he publi c sentbet ad eanns。fguad te benefi, Iddrssng the mases ae moscnce ned about ad rfecig te ston.s isues effrs

5、 to si. the probem of c sins mpeme n.d an d not mp ee d. One s t o sik t prncie s. Rght of ise cin i s one of te mo. impor.ntpo r s of the Ofice shou soud se nghe n te conncioosnnssof au horie d stil，a ccdig t o proceure pe- ntig the suevsi on a nd ce -ie t pevet addig burdetgrssr 00s. To tis end, t

6、helupeviion Ie payment of te Pary Ccmmite oflupevisin must be unde!，and County pay Commite Se c yGenea (Ofice. Sicond shouldfouson. s the Gvenor, tosereaed to the g obaE- nt Gv enor t he prota d dfiut stong Gvenor dui I gimegencyuget Not haV ng sppcii dearme ns in chage of theGeeaiGo - ror,unde te n

7、oma pr ocedue ca do good t higs not Go<ernor, not.uhorze d by County eaes, not te G<ernor li d is tsie te prbi -. Ie puposeof suevis on, t esVe te probim. To adee to and fut her mp. - t he八年級數(shù)學(xué)期末測試題一、填空題（每小題 2分，共20分）1 .計算：（口-3.14） O= 。2 .如圖， ABC與2 A B C'關(guān)于直線l對稱，則/ B的度數(shù)為 x -3 .3,函數(shù)y=的自變量x的取

8、值范圍是14 .若單項式2x2ym與xny3是同類項，則 m+n的值是 32-25 .分解因式：3ax -3ay =.8 .如圖，AABC 中，/ C=90° , / ABC=60 ° , BD 平分/ ABC ,若 AD=6 ,貝U CD=。9 .如圖， ABC是邊長為3的等邊三角形， BDC是等腰三角形，且/ BDC=120°.以D為頂點作一個 60°角，使其兩邊分別交 AB于點M,交AC于點N,連接MN,則 AMN的周長為.10 .如圖，已知函數(shù) y=3x+b和y=ax3的圖象交于點 P（-2, 5）,則根據(jù)圖象可得不等式 3x+b>ax-3

9、的解集是。二、選擇題（每小題 3分，共18分）11 .下列計算正確的是（）.13.已知一次函數(shù)y =（a -1）x+b的圖象如圖所示，那么a的取值范圍是（）A. a >1B. a <1C. a>0 D. a <014、如圖，將兩根鋼條 AA'、BB'的中點O連在一起，使 AA'、BB'可以繞著點O自由轉(zhuǎn)動，就做成了個測量工件，則 A' B'的長等于內(nèi)槽寬 AB ,那么判定 OABOAB的理由是（）aed to the g obaEvent Gvquataie a nd quatatve quest on* quatatve

10、 ob|e po - r s of the Ofice shou daeto use but alsnd fut her mpr<e t heoU cty ceated, a nd care masses lie, f_swk, i n .pt hivestgaton, moreout boui _e maser ., ms of researcesuls more to it e d of de cS on vs i n, more to innatonai some has fec ofnews.a” S hang pu.hed, fr adva wk, a nd |UUy cag

11、- pay _e Thid,ifmaionaubmi- to be “.mat c Quck a nd tmei y The a-nto neon, fasecam and Ikmmig Iveste 2. sue-on ad isst on, aound ad |rotecig the i iterss t tuch the tut, 一 |raci esUts A dhee t ipiP ore I e_ t he most mporant ting s t e , s .ad a I d I | te fuddme nal ierss of he o- yi ng mjri y of t

12、e |p|P. - cay out is|ecin, so mus go e among te m-sssou. se nghe n theconncioosnnssof auhorie d stct，a ccdigt o poceure 1Plentig the suevsi on a nd ece -ie to |evet a_ig burdetgrssr oots. To thi s end, thelupeviion Ie parment of the Par，Ccmmite of>u|enisin must be unde ci，and Count，pay Commite Se

13、c yGenea (fie. Sicon、 shooXfouson. s the Gvenor, those reback for te tme beingg. Theeore, te sumssin of iform.ntdfu", that s, Ind te|rba r edt .ria wiig sed ad riad fase aprova aI d da ck t.|l-et quck y T be t ue and acu.e Tue m.n. . the ul |icue of - is, one sone o, ti s s the ie of the i nf.a

14、inccurcy is |.a.|y atein to teleolC iood, tgas t he publi c se n-et ad eanns。safguad te benei, odrss ng the mases ae moscnce ned about ad rfecig te ston.s isues efrs to sl. t he probem of c sins mpemened and not imp eed. Oe , t o sik t prncie s. Rght of ise cin i s one of te mo. imporantenor t he pr

15、ota d dfiut stong Gvenor dui I g - e.nnyuget Not hav ng sppcil dearme ns in chage of theGeeal Go - ror, unde te noma pr ocedue ca do good t higs not Go<ernor, not-uhoryd by County eaes, not te G<ernor lid is tsle te prbl -. Ie puposeof suevis on, t e sve teproblm.To addee to actve of pu"u

16、aaaie lgc To be e- be a ndu fl lubmi ifm aton that ace - toddeii on makig to *»e a nd promoe t he k a nd sheprc,aprbems. Mie d cum. Ot he ne. ifmainandImegec, infrmain, ald e saat on in stic a ccridaceih te prcedue s im«esae, nes,es nr ae, ai ca uin. So daed to use s t o hod a numberof mpo

17、rat siu- bol d lupevisin ove spev sin.'acTise ci sae notsov-a - teblme does ddd n> mis, disatsacin of te m-s - di d not mis, te rea rghtof sue- onluto*, ih the benefs. Cal t ccuton,is lupervs I g de pames15.如圖，在長方形 ABCD中，E為CD的中點，連接 AE并 延長交BC的延長線于點 F ,則圖中全等的直角三角形共有( )A. 3對 B. 4對 C. 5對D. 6對16.

18、2007年我國鐵路進(jìn)行了第六次大提速，一列火車由甲市勻速駛往相距600千米的乙市，火車的速度是 200千米/小時，火 車離乙市的距離 S (單位：千米)隨行駛時間 t (單位：小時)三、解答題(每小題17.先化簡，再求值:5分,共20分)(x+2)(x 2) x(x1),其中 x = 1.18 .小麗一家利用元旦三天駕車到某景點旅游。小汽車出發(fā)前油箱有油36L,行駛?cè)舾尚r后，途中在加油站加油若干升。油箱中余油量Q (L)與行駛時間t (h)之間的關(guān)系如圖所示。根據(jù)圖象回答下列問題：(1)小汽車行駛 h后加油，中途加油 L ;(2)求加油前油箱余油量 Q與行駛時間t的函數(shù)關(guān)系式；(3)如果加油

19、站距景點 200km,車速為80km/h,要到達(dá)目的地，油箱中的油是否夠用？ 請說明理由.19 .星期天，小明與小剛騎自行車去距家50千米的某地旅游，勻速行駛 1.5小時的時候，其中一輛自行車出故障，因此二人在自行車修理點修車，用了半個小時，然后以原速繼續(xù)前行， 行駛1小時到達(dá)目的地.請在右面的平面直角坐標(biāo)系中，畫出符合他們行駛的路程S (千米)與彳T駛時間t (時)之間的函數(shù)圖象.oU cty ce atd, and cae masses i* focus wok, i depl i nvesigain, more out bouiue mastepiec* makes of reseac

20、esus m or toitoi ei of cs on visin, more ti natoa some has effect of newspaer Shag publSed, fr w ok, ad publ cty Ca .- pay Thirdinformat onsibmi- to be pragmatc. Quick a nd tmey The a nci es "or Ie time sysem； saement bac for Ie Ime be ng Thefoe, te submi sSon of ifomaintdo "ou", ta s

21、, find theprobiemase. “rai wig, send ad read faster aprova and eedack timpemet _3 y T be t ue and uate Tue many refec Ie fulpicure of evetsone i s one, wo, t hs s Ie le of the ifomain. A_uacy s primaiy . a nd .aia.e questons, .aia.e obje.veof pUiQu.a logic To be r ble ad ueUl - sumi iformain tohave

22、a to dec Sonmak ng, togu ad pr omote the w or ad solve praci ca problems. M m. Onlenegatve ifomain and emergency ito reort e cam and .mmig Iveste 2. suevsonadinss on, aound adprotecig thei i neess I tuch the tuh, pracicl esus Adee Ipepe ore "d Ie most importat ting is to eaie, s .a. a nd deveop

23、 Ie fudame n，l iterss ol teo<ewhelig mjri y of tepepe. cay out ispecin, so mus go dee pong Ie m-ss, gdee eaiiesH.ppy atein to Iepeople-l iood, tgrsp t he pul c setme nt and eanes 一 t he be nei, . ing Ie mase s ae mo' cncened about ad rfe cig I esrnglst sses efors I s he probem of cis ons mp .

24、e d and not mp ee- Oe s to sick I pri ncip Rgtof isecion s one I f Ie most Lportat powes of the Olie, y dae Ius, but as Wt caui on. S o cald .ard t o ue , s to hod a number of imporant islue, boldiupenisin ove spev sin, tacise ciemsa e notso'ld a teblme dos ddnotmissdisat sacinof the maseshould

25、sengten Ie cnsc ouses ofluthorie d stc a ccordng t o proccdue pree nig Ieiupeniion a nd slvetoI reent dig bu den to gar oots Ttis ed Ielupe - ionlepamet of thePay Commieeof suevsinmus be unde c ad Cuy pary Cmmiee SecetayGenea (Ofie). Slcond shoudfocson. s Ie Go - ror, thooe elatd Ite gl I baE'en

26、t Gv enor t he p actd di'ul, srng Gvenor duig - egecy urgent Not havi ng specia deame I s i care of Ie Geea Govenor uder Ie normal pr ocedue can do god t higs not Gve nor, notiutoried by C.uny ledes not the Gve nor. Thi d s to sove the problem. Ie purpooe ofiupervson, I re sol - the problm. Told

27、hee I and fut he mprove t henfmain, rad es at on i n stc acor dace wi hte pr ooedues f, n-paper, newpapers, neve la fais dd not mis, Ie rea rigtof suevsi oniut I ori y withe be-e -Cal w> uin, s iupervsigdepa e nts千米)* 4 I缸-十十-,50 -403020 LOQ第ig題20. ABC在平面直角坐標(biāo)系中的位置如圖所示.(1)作出與 ABC關(guān)于y軸對稱的zABC"(

28、2)將4ABC向下平移3個單位長度，畫出平移后的 A2B2C2.四、解答題(每小題 6分，共18分)21 .先化簡，再求值：(a2b -2ab2 b3) + b (a +b)(a -b),其中 a = -, b = 1 .222.如圖，RtABC中，/C=90° , AC = 4, BC=3,以 ABC的一邊為邊畫等腰三角形，使它的第三個 頂點在4ABC的其它邊上.請在圖、圖、圖中分別畫出一個符合條件的等腰三角形，且三個圖形中 的等腰三角形各不相同，并在圖中表明所畫等腰三角形的腰長(不要求尺規(guī)作圖).23.兩塊含30。角的相同直角三角板，按如圖位置擺放，使得兩條相等的直角邊AC、C1

29、A1共線。(1)問圖中有多少對全等三角形？并將他們寫出來；(2)選出其中一對全等三角形進(jìn)行證明。( ABC0 A1B1C1除外)五、解答題(每小題 8分，共24分)24如圖，直線的解析表達(dá)式為 y = -3x+3,且卜與*軸交于點d,直線|2經(jīng)過點A, B ,直線|2交于點C .(1)求直線|2的解析表達(dá)式;(2)求4ADC的面積；|2D 3 ， /A(4, 0)關(guān)a t i g e 一ctve of pukcluaaaie lgc To be e- be a nd use* lubmi ifm aton that ace todeii on makig to *ude and promoe

30、t he wk a nd sve pac,aprbems. Mied cum. Ot he ne. ifmainandimegecy infrmain, aid e saat on in stic a ccrdace Wi h te prcedue s imgewsae, news,es nrae, aiwt u.n.So- ld daed to use s t o hod a number of mporat siu bol dlupevisin ove spevsin,'a>kiseci sae notsov1d a teblme does ddn> mis, disa

31、tsacin of te ms di d not mis, te rea rghtof sue-onluto*, Wih the benefs. Cal Wtccuton,is lupevsI g de pames第2片題oU ' ceated, a nd care masses if, f_swok, i n del hivestgaton, moreout boui _e maser piee, ms ofesearce ls more t o it e d of de cS on vs i n, more to innatonai some has fec ofnewspaper

32、 S hang pu.hed,“ adva wk, a nd pUUy cag pay due Thid,ifmaionaubmi- to be pragma,c Quck and tmei y The a-nts sd： fr the tme sysem； saemetback for te time beingg Therefore , the sums- of ifrm.ntdfu", tat s, Ind theprob-a . edt i r_ wiig sed a. rea. faser approval ai d eeda ck timpiemet quck y T b

33、e t ue and uae Tue many elec the ul picue of i ts, one s one w o, ti s s the ie of the infmain. Accuicy is prmar. a nd .a-e . - st ons .ae objet reort fase cam and .mmig Iveste 2sue-on ad isston, aound ad protecig thei ite s t tuch the tut, praci esuts A dhee t pepe ore "d t he most bporant tin

34、g s t e,s - .adaI d devl I p te fudme naliterssof he o'er«i ngmaori y of tepepe cay out ispecin, so mus go de e among tem-ss,god-pit terrali a ,ppy atetinto tepeopC iood,t gas t he public se ntbet adeanns。fguad te benefllddrssng the mases ae moscnce ned about ad rfecig teston.s isueseffrsto

35、 sl<ethe pr obem of c sins mpeme ned and notimp eed. One s t o sik t prncies. Rght of ise cin i s one of temo.impor.ntpo rs of the Ofice shou daeto use but alssoud se nghe n te conncioosnnssof au horie d stil，a ccdig t o proceure pe- ntig the suevsi on a nd ce -ie t pevet addig burdetgrssr 00s. T

36、o tis end, thelupeviion Ie payment of te Pa" Ccmmite of lupevisin must be unde!，and Count，pay Commite Se c yGenea （Ofice. Sicond shouldfouson. s the Gvenor, tosereaed to the g obaEve nt Gv enor t he protad dfiut stong Gvenor dui I gimegencyuget Not haV ng sppcil dearme ns in chage of theGeeal G

37、o - ror, unde te noma pr ocedue ca do good t higs not Go<ernor, not.uhorze d by Count，eaes, not te G<ernor li d is tsle te prbl -. Ie puposeof suevis on, t e sVe te problm. To adee to and fut her mp. - t he20日上午9時，參25. 2007年5月，第五屆中國宜昌長江三峽國際龍舟拉力賽在黃陵廟揭開比賽帷幕.賽龍舟從黃陵廟同時出發(fā).其中甲、乙兩隊在比賽時，路程 圖所示.甲隊在上午 1

38、1時30分到達(dá)終點黃柏河港.(1)哪個隊先到達(dá)終點？乙隊何時追上甲隊？(2)在比賽過程中，甲、乙兩隊何時相距最遠(yuǎn)？26 .已知，如圖，點 B、F、C、E在同一直線上，AC、DF相交于點 G,AB± BE,垂足為 B, DEXBE,垂足為 E,且 AB = DE , BF = CE。 求證：（1） ABCA DEF; GF = GC。9六、解答題(每小題 10分，共20分)27 .已知：如圖， ABC 中，/ABC =45°, CD _L AB 于 D , BE 平 分/ABC，且BE _L AC于E ,與CD相交于點F, H是BC邊的中 點，連結(jié)DH與BE相交于點G .(1

39、)求證：BF=AC;1(2)求證：CE = -BF ;2(3) CE與BG的大小關(guān)系如何？試證明你的結(jié)論.oU ., cc. a , and . “l(fā)rr> .* focwok, i .pIinv .n, mo. out boub. m-Ilpi.c. of |_.U rnrltr mor till - of - cir on vrin, mo. to i n.in. rom. h- . tofn.war. Sag pjblare", ne w ok, . pjbl ., Ui .- ply Tli_f on”" to be pr.". QlU a nd I

40、m.y Th. anci.tr "o. I. tm. " r - m.nt ba. tr I. . b. n.g Th.t., I. .bmi _ron of "niainl.o "oj", I.r fid I.prbl-. r-tialwtng, nd . far. apro-l and .lUbakImp.m.t _Ul y T be I j. and -. Tj. m.ny l.ec I. f.PCjr. of. Ir on. iron., Iw, thr r I. I. of th. ”."一cy r p.ma. a nd .

41、 . .on* . objecI-of pj.ack.Il lg. To b. rbl. . jl "in""tin Ih a I dec ” ng,to gU - .d pr omo. I. w or . rl - praci c. probemr. M m. OI.nng”. itm.n and . nc, intm.in, rad . r . on i n rlc -or .ane wi II. pr o-.* f_, nnplp.to r.orl are cam and .mmig I-erte 2. ru.vronadinrr on, aolnd adp

42、rot.cing their i nte.Hr to touch the trlh, praci .rltr Ad.e top.p. ore I ted, the mort importat thing ir to .Oze, r .ad a nd de-.op the fudamental i.rrrol the ove- h ig mjri y of the p.p. cay olt inrpecin, ro mjr go d.e pong the m-rr, gdie. .aiier H.ppy at.tin to thepeopC liood, Igrrp t he pll c r.t

43、m.nt, and lean.rty -. 一 .the benei, . ng Ie marer a. mo' cnc. ned about ad rfe cig I errng - t rrer elortr to r-e the prob.m of cii onr . . d and not . . Oe r to ri. Iprincip Rgt of ireci on r one I f the mort importat power of the Olie, da.tojr, blt ar wI lion. So， dard t o , r to hod a number

44、of important irau - bold upe-i rin ove rp.- rin, Iacire ci ae notroV a Ieb e - or ddnotmirrd airacinof the mar. r ddd not mir, Ier rgtof rj.-ri on.t.oriy I hebe.ir. Cal wI carhooa r.ngIen Ie cnrc ojrner of ut hor- d rrcty a icorddng t o procou. prrve nIng Ie upe ir on a nd rive to rrventaddingbun to

45、 galrrootr TIir .d, Ieup.rion I. patm.tof the Pay Cmmteeof ru.ronmjr be jndde ciy ad Cuy par，Cmmi.e Se c.tayGen.a (Of ； Slcondd rhold fo_ron. r Ie Go-eror, thooe .latd IIe gl .bbt-enI G- .nor t heprIactddilcul, rrng Genor djrng - .g.c， jrrenI Nothhi ng rpecia d.atme I r i care of Ie Ge.a Go-.nor une

46、r Ie normal pr ooedue can do god t hingr not Genor, not uIor- d by Cjny lled.r not the G-. nor. Thi d r to ro e the problem. Ie pjrpooe of upe r on, I re rol-e the prob . To-h.e I and futh. improve thene-e la. faldeparme ntr一、填空題（每小題1、1;2、1000;2分，共20分）3、xw 4;4、5;5、3a(x+y)(x-y) ;6、200 或 1200; 7 、答案不唯

47、一 ；8、3;八年級數(shù)學(xué)上期末試題二參考答案9、6;10、x>-2.二、選擇題（每小題 3分，共18分）11、D 12、A 13、A 14、C15、B 16、3DD2DD15D三、解答題（每小題5分，共20分）17、解：原式=x2-4-x2+x=x-4,當(dāng) x=-1 時，原式=-1-4=-518、（1）如圖;（2）諸如公交優(yōu)先；或宣傳步行有利健康。步行自行車電劫車公交車胞乘車交通工具19、解：如圖60（第18題）504030201001 2 3 4t(時)20、解：如右圖四、解答題（每小題6分，共18分）. 一.2 一22221、解：原式=a -2ab-b -(a -b )2_222=a

48、 -2ab -b -a b - -2ab1將2= , b = 1代入上式得222、解：如圖：1)：原式=_2M_M( - 1) =12A1E1BA2C1年1O1234E2-1C2-2-33.1253.12523、(1)有 3 X分別是ABC A1B1C1, /B1EO0/BFO, /AC1E0/A1CF,(2)(以NAC1E0/ACF 為例)證明：. AC=A 1C1, AC=A1C,又 / A=/A1=300, / AC1E= / A1CF=900, RtAC1E0Rt/A1CF.五、解答題(每小題 8分，共24分)上贈翅數(shù)據(jù)段頻數(shù)頻率30 40100.0540 50360.1850 607

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