2020年包頭市初二數(shù)學上期中試題(帶答案)

A.110°/r/nA.110°/r/nB.120"/r/nC.125°/r/nD.135°/r/n2020/r/n年包頭市初二數(shù)學上期中試題（帶答案）/r/n一、選擇題/r/n1./r/n己知一個正多邊形的內(nèi)角是/r/n140°/r/n,則這個正多邊形的邊數(shù)是（）/r/nA./r/n9/r/nB./r/n8/r/nC./r/n7/r/nD./r/n6/r/n2./r/n如圖，在/r/naAEC/r/n中，/r/nED/r/n平分/r/nZABC,EC/r/n的垂直平分線交/r/nBD/r/n于點/r/nE,/r/n連接/r/nCE,/r/n若/r/nZA=60°,ZACE=24°,/r/n則/r/nZABE/r/n的度數(shù)為（）/r/nA/r/n?士/r/nB./r/n竺/r/nx/r/n+/r/ny/r/n4ci/r/nC/r/n?/r/n32°/r/n)/r/nx/r/n2/r/n-4/r/nC./r/nx-2/r/nD.48°/r/na/r/n2/r/n+4a/r/nD./r/n4./r/n從甲地到乙地有兩條路：一條是全長/r/n750km/r/n的普通公路,/r/n另一條是全長/r/n600km/r/n高速公/r/n路?某客車從甲地出發(fā)去乙地，若走高速公路，則平均速度是走普通公路的平均速度的/r/n2/r/n倍，所需時間比走普通公路所需時間少/r/n5/r/n小時?設(shè)客車在普通公路上行駛的平均速度是/r/nxkm/h,/r/n則下列等式正確的是（）/r/n600/r/n750/r/n600/r/n750/r/nA./r/n——+5=/r/nB./r/n——/r/n-5=/r/nX/r/n2x/r/nX/r/n2x/r/n600/r/n750/r/n600/r/n750/r/nC./r/n+5=/r/nD./r/n——/r/n-5=/r/n2x/r/nX/r/n2x/r/nX/r/n乩/r/n卞歹/r/nll/r/n條件巾能/r/n$teAABC^ADEF/r/n的果/r/n()/r/nA/r/n?/r/nAB=DE,/r/nEC=EF,ZA=ZD/r/nB.ZA=ZD,/r/nZB=ZE,ZC=ZF/r/nC.AC=DF,/r/nZB=ZF,AB=DE/r/nD/r/n?/r/nZB=ZE,/r/nZC=ZF,AC=DF/r/n6/r/n.計算帥-①亍的結(jié)果為（）/r/nA./r/n丄/r/nB./r/nC.-x/r/n2/r/ny/r/nD./r/n一小/r/nV/r/n7./r/n如圖/r/n、/r/nABHCD/r/n、/r/nDE/r/n丄/r/nBE,BF/r/n、/r/nDF/r/n分別為/r/nZABE.ACDE/r/n的角平分線，/r/n()/r/n如圖，把三角形紙片/r/nABC/r/n沿/r/nDE/r/n折疊，當點/r/nA/r/n落在四邊形/r/nBCDE/r/n外部時，則/r/nZA/r/n與/r/nC.3ZA=2Z1-Z2/r/nZl.Z2/r/n之間的數(shù)量關(guān)系是(/r/nC.3ZA=2Z1-Z2/r/nB.3ZA=2(Z1-Z2)/r/nD/r/n?/r/nZA=Z1-Z2/r/n己知/r/nx/r/n2/r/n-4x-l=0＞/r/n則代數(shù)式/r/n2x(x—3)—(x—/r/n1/r/n尸+/r/n3/r/n的值為()/r/nTOC\o"1-5"\h\z/r/nA./r/n3/r/nB./r/n2/r/nC./r/n1/r/nD.-1/r/n已知/r/nx+y=5,xv=6,/r/n則的值是/r/n(/r/n)/r/nA.1B/r/n?/r/n13C/r/n?/r/n17D/r/n?/r/n25/r/n1/r/n2/r/n3/r/n11/r/n-/r/n式子：喬/r/n？/r/n正'莎的最簡公分母是/r/n(/r/n)/r/nD.a/r/n8/r/n-b/r/ns/r/nA./r/n24x/r/n2/r/ny/r/n2/r/nxy/r/nB./r/n24/r/nx/r/n2/r/ny/r/n2/r/nC.12F/r/nD.a/r/n8/r/n-b/r/ns/r/n12./r/n計算：/r/n(a/r/n—/r/nb)(a/r/n+/r/nb)(a/r/n2/r/n+/r/nb/r/n2/r/n)(a/r/n4/r/n—/r/nb/r/n4/r/n)/r/n的結(jié)果是(/r/n)/r/nA.a/r/ns/r/n+2a/r/n4/r/nb/r/n4/r/n+b/r/ns/r/nB.a/r/ns/r/n-2a/r/n4/r/nb/r/n4/r/n+b/r/ns/r/nC.a/r/ns/r/n+b/r/ns/r/n二填空題/r/n13/r/n?/r/n已知//r/nb,c/r/n是/r/nZkABC/r/n的三邊長，/r/na,b/r/n滿足/r/n|a-7|+(b-1)/r/n2/r/n=0,c/r/n為奇數(shù)，則/r/nc=/r/n?/r/n14./r/n如圖所示，過正五邊形/r/nA3CDE/r/n的頂點/r/n3/r/n作一條射線與其內(nèi)角/r/nZE43/r/n的角平分線相交/r/n于點/r/nP/r/n，/r/n且/r/nZABP=60°,/r/n則/r/n度./r/n15/r/n?/r/n已知射線/r/nOM/r/n?以/r/nO/r/n為圓心，任意長為半徑畫弧，與射線/r/nOM/r/n交于點/r/nA,/r/n再以點/r/nA/r/n為圓/r/n兩弧交于點/r/nE,/r/n畫射線/r/nOE,/r/n如圖所示,/r/n則/r/n兩弧交于點/r/nE,/r/n畫射線/r/nOE,/r/n如圖所示,/r/n則/r/nZAOB=/r/n(度)/r/nY—/r/n2/r/n16/r/n-/r/n當時’分式后的值為零./r/n一個等腰三角形的兩邊長分別為/r/n4cm/r/n和/r/n9cm,/r/n則它的周長為/r/n_cm/r/n?/r/n若關(guān)于/r/nx/r/n的方程/r/n—/r/n—無解，則仆/r/nx_5/r/n10-2x/r/n關(guān)于/r/nX/r/n的分式方程上$=/r/n1/r/n的解為負數(shù)，則/r/nd/r/n的取值范圍是/r/n?/r/nX+1/r/n如果一個正多邊形每一個內(nèi)角都等于/r/n144/r/n。，那么這個正多邊形的邊數(shù)是—/r/n./r/n三、解答題/r/n說明代數(shù)式/r/n[U-y)/r/n2/r/n-U+y)U-y)]-(-2/r/n)/r/n-)+/r/n的值，與/r/ny/r/n的值無關(guān)./r/n解分式方程：/r/n?/r/nx+2/r/nJT/r/n一/r/n4/r/n如圖，在/r/nAABC/r/n和/r/nZkAED/r/n中，/r/nAC/r/n與/r/nED/r/n相交于點/r/nE,AD=BC,ZDAB=ZCBA,/r/n求/r/niiE/r/n：/r/nAC=BD/r/n?/r/n為了對學生進行革命傳統(tǒng)教育，紅旗中學開展了“清明節(jié)祭掃”活動.全校學生從學校同時出發(fā)，步行/r/n4000/r/n米到達烈士紀念館.學校要求九⑴班提前到達目的地，做好活動的準備工作.行走過程中，九/r/n(/r/n1)/r/n班步行的平均速度是其他班的/r/n1.25/r/n倍，結(jié)果比其他班提前/r/n10/r/n分鐘到達.分別求九/r/n(/r/n1/r/n〉班、其他班步行的平均速度./r/n已知/r/na=2+JL/r/nb=2/r/n-*/r/n求下列各式的值：/r/n(1)/r/na/r/n2/r/n+2ab+b/r/n2/r/n(2)/r/ncr-b/r/n2/r/n【參考答案】杯*試卷處理標記,請不要刪除/r/n選擇題/r/n1/r/n./r/nA/r/n解析：/r/nA/r/n【解析】/r/n分析：根據(jù)多邊形的內(nèi)角和公式計算即可./r/n詳解：/r/n?V-/r/n1/r/nW)/r/n=360°/r/n-T/r/n40°/r/n=9./r/n答:這個正多邊形的邊數(shù)是/r/n9/r/n.故選/r/nA./r/n點睛：本題考查了多邊形，熟練掌握多邊形的內(nèi)角和公式是解答本題的關(guān)鍵./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/n先根據(jù)/r/nEC/r/n的垂直平分線交/r/nBD/r/n于點/r/nE/r/n證明/r/naBFE^aCFE/r/n(SAS)/r/n,/r/n根據(jù)全等三角形的性質(zhì)和角平分線的性質(zhì)得到/r/nZA/r/n恥二/r/n=/r/n,再根據(jù)三角形內(nèi)角和定理即可得到/r/n答案./r/n【詳解】/r/n解：如圖：/r/nVBC/r/n的垂直平分線交/r/nBD/r/n于點/r/nE,/r/n.??BF/r/n二/r/nCF,ZBFE=ZCFE=90°,/r/nffiABFE/r/n和/r/nZkCFE/r/n中，/r/nEF/r/n=/r/nEF/r/n</r/nZEFB/r/n=/r/nZEFC/r/nBF=CF/r/nAaBFE^aCFE/r/n(SAS),/r/n:/r/n.ZEBF=ZECF/r/n(全等三角形對應(yīng)角相等)，/r/n又/r/nVBD/r/n平分/r/nZABC,/r/n?/r/n?/r/n?/r/nZABE/r/n=/r/nAEBF/r/n=/r/nZECF/r/n,/r/n又/r/nV/r/nZABE+ZEBF/r/n+/r/nZECF/r/n+/r/nZACE+/r/nZA/r/n=180°/r/n(三角形內(nèi)角和定理)，/r/n?/r/n?/r/n?/r/nZABE/r/n+/r/nAEBF/r/n+/r/nZECF/r/n=180°-60°-24°=96°,/r/n?/r/n?/r/n?/r/nZABE=-x96°=32°/r/n,/r/n3/r/n故選/r/nC./r/n【點睛】/r/n本題主要考查了三角形全等的判定與性質(zhì)、角平分線的性質(zhì)、三角形內(nèi)角和定理，證明/r/nZABE/r/n=/r/nZEBF/r/n=/r/nZECF/r/n是解題的關(guān)鍵./r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n根據(jù)最簡分式的定義：分子和分母中不含公分母的分式，叫做最簡分式，對四個選項中的/r/n

/r/n分式一一判斷即可得出答案./r/n【詳解】/r/n/+卩/r/n2/r/n解：/r/nA/r/n——/r/n,分式的分子與分母不含公因式，是最簡分式；/r/nx/r/n+/r/ny/r/n6h3h/r/nB.—=—,/r/n分式的分子與分母含公因式/r/n2,/r/n不是最簡分式；/r/n4a/r/n2a/r/n尢/r/n2—4/r/nC/r/n+2,/r/n分式的分子與分母含公因式心/r/n2,/r/n不是最簡分式；/r/nX-/r/n2/r/n(/r/n1^+/r/n4(/./r/nD/r/n=q+4,/r/n分式的分子與分母含公因式/r/n4/r/n不是最簡分式，/r/na/r/n故選/r/nA./r/n【點睛】/r/n本題考查了最簡分式的概念?對每個分式的分子和分母分別進行因式分解是解題的關(guān)鍵./r/n4/r/n./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/n分別表示出客車在普通公路和高速公路上行駛的時間，即可得到方程./r/n【詳解】/r/n根據(jù)題意：客車在普通公路上行駛的時間是竺小時，在高速公路上行駛的時間是空/r/nx/r/n2x/r/n小時，由所需時間比走普通公路所需時間少/r/n5/r/n小時可列方程：竽/r/n+5=—,/r/n2x/r/nx/r/n故選：/r/nC./r/n【點睛】/r/n此題考查分式方程的實際應(yīng)用，正確理解題意找到等量關(guān)系是解題的關(guān)鍵./r/nD/r/n解析：/r/nD/r/n【解析】/r/n分析：根據(jù)全等三角形的判定定理/r/nAAS,/r/n可知應(yīng)選/r/nD./r/n詳解：解：如圖:/r/n詳解：解：如圖:/r/n4/r/n選項中根據(jù)/r/nAB=DE,BC=EF/r/n,/r/n4=ND/r/n不能判定兩個三角形全等，故/r/nA/r/n錯：/r/nB/r/n選項三個角相等，不能判定兩個三角形全等，故/r/nE/r/n錯；/r/nC/r/n選項看似可用“邊角邊”定理判定兩三角形全等，而對照圖形可發(fā)現(xiàn)它們并不符合此判定條件，故/r/nC/r/n錯；/r/nD/r/n選項中根據(jù)/r/n“4AS”/r/n可判定兩個三角形全等，故選/r/nD/r/n：/r/n點睛：本題考查了全等三角形的條件，本題沒有給出圖形，增加此題的難度.若能順利畫出圖形，對照圖形和選項即可得到正確選項./r/nC/r/n解析:/r/nC/r/n【解析】/r/n【分析】/r/n根據(jù)分式的減法和除法可以解答本題/r/n【詳解】/r/n=-x(x-y)-/r/nx-y/r/n=-x/r/n2/r/ny/r/n故答案為/r/nc/r/n【點睛】/r/n本題考查分式的混合運算，解答本題的關(guān)鍵是明確分式混合運算的計算方法./r/nD/r/n解析：/r/nD/r/n【解析】/r/n【分析】/r/n【詳解】/r/n如圖所示，過￡作/r/n￡G/r/n〃/r/nAB./r/n?:/r/nAB//CD,/r/n:/r/n.EG//CD/r/n,/r/nZABE+ZBEG=180/r/n。，/r/nZCDE/r/n十/r/nZDEG=/r/n180°,/r/n?/r/n?.ZABE/r/n十/r/nZBED/r/n十/r/nZCDE=360/r/n。/r/n./r/n又/r/n?；/r/nDE/r/n丄/r/nBE,BF,DF/r/n分別為/r/nZABE,ZCDE/r/n的角平分線，/r/n11/r/n:/r/n.ZFBE+ZFDE=—/r/n(/r/n上/r/nABE+ZCDE)=—/r/n(360/r/n。/r/n-/r/n90/r/n。)/r/n=135/r/n。，/r/n./r/n?/r/n.ZBFD=360°/r/n-/r/nZFBE/r/n-/r/nZFDE/r/n-/r/nZBED=360/r/n。/r/n-135°-90°=135°./r/n故選/r/nD./r/n【點睛】/r/n本題主要考查了平行線的性質(zhì)以及角平分線的定義的運用，解題時注意：兩直線平行，同/r/n旁內(nèi)角互補.解決問題的關(guān)鍵是作平行線./r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n根據(jù)折疊的性質(zhì)可得/r/nZAJZA,/r/n根據(jù)平角等于/r/n180/r/n。用/r/nZ1/r/n表示出/r/nZADA\/r/n根據(jù)三角形的一個外角等于與它不相鄰的兩個內(nèi)角的和，用/r/nZ2/r/n與/r/nZA/r/n，/r/n表示出/r/nZ3,/r/n然后利用三角形的內(nèi)角和等于/r/n180/r/n。列式整理即可得解./r/n【詳解】/r/n如圖所示:/r/n?/r/n?/r/n?AA^DE/r/n是/r/n△/r/nADE/r/n沿/r/nDE/r/n折疊得到，/r/n:./r/nZA*=ZA,/r/n又/r/nV/r/nZADA/r/nr/r/n=180°-Zl,Z3=ZA'+Z2,/r/nTZA+ZADA/r/nx/r/n+Z3=180/r/no/r/n,/r/n即/r/nZA+180°-Zl+ZA/r/n,/r/n+Z2=180°,/r/n整理得，/r/n2ZA=Z1-Z2./r/n故選/r/nA./r/n【點睛】/r/n考查了三角形的內(nèi)角和定理以及折疊的性質(zhì)，根據(jù)折疊的性質(zhì)，平角的定義以及三角形的一個外角等于與它不相鄰的兩個內(nèi)角的和的性質(zhì)，把/r/nZl/r/n、/r/nZ2/r/n、/r/nZA/r/n轉(zhuǎn)化到同一個三角形中是解題的關(guān)鍵./r/n9./r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n先將原代數(shù)式進行去括號化簡得出/r/nF/r/n-4/+2/r/n，/r/n然后根據(jù)/r/nx/r/n2/r/n-4x-l=0/r/n得出/r/nx/r/n2/r/n-4x=l/r/nf/r/n最后代入計算即可./r/n【詳解】/r/n由題意得：/r/n2x(x—3)—(x—1)'+3=F—4x+2/r/n，/r/n*/x/r/n2/r/n-4x-l=/r/n0?/r/n/.x/r/n2/r/n-4x=ly/r/n?I/r/n原式/r/n=/r/n-/r/n4x+2=1/r/n+2=3/r/n./r/n故選：/r/nA./r/n【點睛】/r/n本題主要考查了整式的化簡求值，整體代入是解題關(guān)鍵./r/n10/r/n./r/nB/r/n解析：/r/nB/r/n【解析】/r/n【分析】/r/n將/r/nx+y=5/r/n兩邊平方，利用完全平方公式化簡，把/r/nxy/r/n的值代入計算，即可求出所求式子的值./r/n【詳解】/r/n解/r/n：/r/n將/r/nx+y=5/r/n兩邊平方得：/r/n（/r/nx+y）/r/n2/r/n=x/r/n2/r/n+2xy+y/r/n2/r/n=2/r/n5,/r/n將/r/nxv=6/r/n代入得：/r/nx/r/n2/r/n+12+y/r/n2/r/n=25,/r/n則/r/n故選：/r/nB./r/n【點睛】/r/n此題考查了完全平方公式，熟練掌握完全平方公式是解本題的關(guān)鍵./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/n分母都是單項式，根據(jù)最簡公分母的求法：系數(shù)取最大系數(shù)，不同字母取最高次幕，將它們相乘即可求得./r/n【詳解】/r/n1/r/n2/r/n3/r/n式子：的最簡公分母是：/r/n12F/r/n）/r/nd/r/n2x~y3x~4xy~/r/n故選：/r/nC./r/n【點睛】/r/n本題考查最簡公分母的定義與求法./r/nD/r/n解析：/r/nD/r/n【解析】/r/n試題分析：根據(jù)平方差公式可直接求解，即原式=/r/n（/r/na/r/n2/r/n-b/r/n2/r/n）（/r/na/r/n2/r/n+b/r/n2/r/n）（/r/na/r/n4/r/n+b/r/n4/r/n）/r/n=/r/n（/r/na/r/n4/r/n-b/r/n4/r/n）/r/n+/r/n故選/r/nD/r/n考點：平方差公式/r/n二、填空題/r/n7/r/n【解析】【分析】根據(jù)非負數(shù)的性質(zhì)列式求出/r/nab/r/n的值再根據(jù)三角形的任意兩邊之和大于第三邊兩邊之差小于第三邊求出/r/nc/r/n的取值范圍再根據(jù)/r/nc/r/n是奇數(shù)求出/r/nc/r/n的值【詳解】/r/nTab/r/n滿足/r/n|a/r/n-7|+/r/n（b-/r/n1）/r/n2=0/.a/r/n-7/r/n解析：/r/n7/r/n【解析】/r/n【分析】/r/n根據(jù)非負數(shù)的性質(zhì)列式求出/r/na/r/n、/r/nb/r/n的值，再根據(jù)三角形的任意兩邊之和大于第三邊，兩邊之差小于第三邊求出/r/nc/r/n的取值范圍，再根據(jù)/r/nc/r/n是奇數(shù)求出/r/nc/r/n的值./r/n【詳解】/r/nVa,b/r/n滿足/r/n|a/r/n?/r/n7|+(b-/r/n1)/r/n2/r/n=0,/r/n/.a-7=0,b-1=0,/r/n解得/r/na=7,b=l,/r/nV7/r/n-/r/n1=6,7+1=8,/r/n?/r/n°/r/n?/r/n6vcv8,/r/n又/r/nTc/r/n為奇數(shù)，/r/n??/r/n?/r/nc=7,/r/n故答案為/r/n7./r/n【點睛】/r/n本題考查非負數(shù)的性質(zhì)：偶次方，解題的關(guān)鍵是明確題意，明確三角形三邊的關(guān)系./r/n66/r/n［/r/n解析】【分析】首先根據(jù)正五邊形的性質(zhì)得到度然后根據(jù)角平分線的定義得到度再利用三角形內(nèi)角和定理得到的度數(shù)【詳解】解：/r/n??/r/n?五邊形為正五邊形.??度/r/n??/r/n?是的角平分線.??度/r/n??????/r/n故答案為：/r/n66/r/n【點睛】本題考查了多/r/n解析：/r/n66/r/n【解析】/r/n【分析】/r/n首先根據(jù)正五邊形的性質(zhì)得到/r/n=/r/n度，然后根據(jù)角平分線的定義得到/r/nZPAB/r/n=54/r/n度，再利用三角形內(nèi)角和定理得到/r/nZAPB/r/n的度數(shù)./r/n【詳解】/r/n解：/r/n??/r/n?五邊形/r/nA8CQE/r/n為正五邊形，/r/n?/r/n?/r/n?/r/nZE4B=108/r/n度，/r/n?/r/n?/r/n?AP/r/n是/r/nZE43/r/n的角平分線，/r/n/.ZPAB/r/n=/r/n54/r/n度，/r/n?/r/n?/r/n?ZABP=6O°,/r/n?/r/n?/r/n?/r/nZAPB/r/n=180°-/r/n60°-54°=66°/r/n?/r/n故答案為：/r/n66./r/n【點睛】/r/n本題考查了多邊形內(nèi)角與外角，題目中還用到了角平分線的定義及三角形內(nèi)角和定理./r/n60/r/n【解析】【分析】首先連接/r/nAB/r/n由題意易證得/r/nAAOB/r/n是等邊三角形根據(jù)等邊三角形的性質(zhì)可求得/r/nZAOB/r/n的度數(shù)【詳解】連接/r/nAB/r/n根據(jù)題意得：/r/n0B/r/n二/r/n0A/r/n二/r/nAB/r/n?/r/n?/r/n?△/r/nAOB/r/n是等邊三角形二/r/nZAOB/r/n二/r/n60°/r/n故答案為：/r/n解析：/r/n60/r/n【解析】/r/n【分析】/r/n首先連接/r/nAB,/r/n由題意易證得/r/nAAOB/r/n是等邊三角形，根據(jù)等邊三角形的性質(zhì)，可求得/r/nZAOB/r/n的度數(shù)./r/n【詳解】/r/n連接/r/nAB,/r/n根據(jù)題意得：/r/nOB=OA=AB,/r/n:./r/n/\AOB/r/n是等邊三角形，/r/n/.ZA(95=60°./r/n【點睛】/r/n本題考查了等邊三角形的判定與性質(zhì).此題難度不人，解題的關(guān)鍵是能根據(jù)題意得到/r/nOB=OA=AB./r/n16./r/n2/r/n【解析】由題意得：解得：/r/nx=2/r/n故答案為/r/n2/r/n解析：/r/n2/r/n【解析】/r/nx—/r/n2=0/r/n由題意得：/r/n｛cC/r/n，解得：/r/nx=2/r/n.故答案為/r/n2/r/nx+2h/r/n0/r/n17/r/n?/r/n22/r/n【解析】【分析】底邊可能是/r/n4/r/n也可能是/r/n9/r/n分類討論去掉不合條件的然后可求周長【詳解】試題解析：①當腰是/r/n4cm/r/n底邊是/r/n9cm/r/n時：不滿足三角形的三邊關(guān)系因此舍去②當?shù)走吺?r/n4cm/r/n腰長是/r/n9cm/r/n時能構(gòu)成三角形則/r/n解析：/r/n22/r/n【解析】/r/n【分析】/r/n底邊可能是/r/n4,/r/n也可能是/r/n9,/r/n分類討論，去掉不合條件的，然后可求周長./r/n【詳解】/r/n試題解析：①當腰是/r/n4cm,/r/n底邊是/r/n9cm/r/n時：不滿足三角形的三邊關(guān)系，因此舍去./r/n②當?shù)走吺?r/n4cm,/r/n腰長是/r/n9cm/r/n時，能構(gòu)成三角形，則其周長/r/n=4+9+9=22cm./r/n故填/r/n22./r/n【點睛】/r/n本題考查了等腰三角形的性質(zhì)和三角形的三邊關(guān)系：已知沒有明確腰和底邊的題目一定要想到兩種情況，分類進行討論，還應(yīng)驗證各種情況是否能構(gòu)成三角形進行解答./r/n18/r/n?/r/n-8/r/n【解析】【分析】試題分析：???關(guān)于/r/nx/r/n的方程無解/r/n???/r/nx/r/n二/r/n5/r/n將分式方程去/r/n分母得/r/n:/r/n將/r/nx/r/n二/r/n5/r/n代入得：/r/nm=-8［/r/n詳解】請在此輸入詳解！/r/n解析：/r/n?/r/n8/r/n【解析】/r/n【分析】/r/n試題分析：/r/n??/r/n?關(guān)于/r/nx/r/n的方程口=」一無解,/r/nAx=5/r/nx-5/r/n10-2x/r/n將分式方程口/r/n■UTT/r/n；/r/n出「去分母得：/r/n2/r/n（/r/nx—l）=—m,/r/nx_>/r/n10-2x/r/n'/r/n7/r/n將/r/nx=5/r/n代入得：/r/nm=/r/n-8/r/n【詳解】/r/n請在此輸入詳解！/r/n19./r/n【解析】【分析】分式方程去分母轉(zhuǎn)化為整式方程由分式方程的解為負數(shù)求出/r/na/r/n的范圍即可【詳解】分式方程去分母得/r/n：/r/n2x+a=x+l/r/n解得/r/n:x=l-a/r/n由分式方程解為負數(shù)得到/r/nl-a<0/r/n且/r/n1-/r/n曲-/r/n1/r/n解得:/r/na>l/r/n且/r/n解析：/r/na/r/n>1/r/n且°/r/nH/r/n2/r/n【解析】/r/n【分析】/r/n分式方程去分母轉(zhuǎn)化為整式方程，由分式方程的解為負數(shù)，求出/r/na/r/n的范/r/nI/r/n制即可/r/n【詳解】/r/n分式方程去分母得/r/n：/r/n2x+a=x+l/r/n解得/r/n:x=l-a,/r/n由分式方程解為負數(shù)，得到/r/nM<0,/r/n且/r/n1-a^-l/r/n解得:/r/na>l/r/n且/r/na=2,/r/n故答案為:/r/na>l/r/n且/r/naH2/r/n【點睛】/r/n此題考查分式方程的解，解題關(guān)鍵在于求出/r/nx/r/n的值再進行分析/r/n20/r/n?/r/n10/r/n【解析】【分析】設(shè)正多邊形的邊數(shù)為/r/nn/r/n然后根據(jù)多邊形的內(nèi)角和公式列方程求解即可【詳解】解：設(shè)正多邊形的邊數(shù)為/r/nn/r/n由題意得二/r/n144°/r/n解得/r/n210/r/n故答案為/r/n10/r/n【點睛】本題考查了多邊形的內(nèi)角與外角熟記公式/r/n解析：/r/n10/r/n【解析】/r/n【分析】/r/n設(shè)正多邊形的邊數(shù)為/r/n11,/r/n然后根據(jù)多邊形的內(nèi)角和公式列方程求解即可./r/n【詳解】/r/n解：設(shè)正多邊形的邊數(shù)為/r/nm/r/n由題意得，/r/n（〃_/r/n2）?180/r/n=]44/r/n。,/r/nn/r/n解得/r/nn=10./r/n故答案為/r/n10./r/n【點睛】/r/n本題考查了多邊形的內(nèi)角與外角，熟記公式并準確列出方程是解題的關(guān)鍵./r/n三、解答題/r/n說明見解析./r/n【解析】/r/n試題分析：根據(jù)整式的混合運算的法則和順序，先算完全平方和平方差，然后合并同類項化簡，通過關(guān)化簡可判斷./r/n試題解析：原式/r/n=(x/r/n2/r/n-/r/n2^/r/n4-y/r/n2/r/n-A-/r/n2/r/n+/r/ny/r/n2/r/n)/r/n(-2y/r/n)/r/n4/r/n-/r/ny/r/n=x-v+v/r/n???/r/n代數(shù)式的值與/r/ny/r/n無關(guān)./r/n原方程無解/r/n【解析】/r/n【分析】/r/n先找出方程的最簡公分母，然后方程兩邊的每一項去乘最簡公分母，化為整式方程，再求解，注意分式方程要檢驗./r/n【詳解】/r/n方程兩邊同乘以/r/n(/r/nx+2)(x-2)/r/n得：/r/n(x-2)/r/n2/r/n一/r/n(x+2)/r/n(x-2)/r/n=16,/r/n解得/r/n：/r/nx=-2,/r/n檢驗：當/r/nx=-2/r/n時，/r/n(/r/nx+2)(x-2)/r/n=0,/r/n所以/r/