2020年包头市初二数学上期中试题(带答案)

A.110°/r/nA.110°/r/nB.120"/r/nC.125°/r/nD.135°/r/n2020/r/n年包头市初二数学上期中试题（带答案）/r/n一、选择题/r/n1./r/n己知一个正多边形的内角是/r/n140°/r/n,则这个正多边形的边数是（）/r/nA./r/n9/r/nB./r/n8/r/nC./r/n7/r/nD./r/n6/r/n2./r/n如图，在/r/naAEC/r/n中，/r/nED/r/n平分/r/nZABC,EC/r/n的垂直平分线交/r/nBD/r/n于点/r/nE,/r/n连接/r/nCE,/r/n若/r/nZA=60°,ZACE=24°,/r/n则/r/nZABE/r/n的度数为（）/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从甲地到乙地有两条路：一条是全长/r/n750km/r/n的普通公路,/r/n另一条是全长/r/n600km/r/n高速公/r/n路•某客车从甲地出发去乙地，若走高速公路，则平均速度是走普通公路的平均速度的/r/n2/r/n倍，所需时间比走普通公路所需时间少/r/n5/r/n小时•设客车在普通公路上行驶的平均速度是/r/nxkm/h,/r/n则下列等式正确的是（）/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条件巾能/r/n$teAABC^ADEF/r/n的果/r/n()/r/nA/r/n・/r/nAB=DE,/r/nEC=EF,ZA=ZD/r/nB.ZA=ZD,/r/nZB=ZE,ZC=ZF/r/nC.AC=DF,/r/nZB=ZF,AB=DE/r/nD/r/n・/r/nZB=ZE,/r/nZC=ZF,AC=DF/r/n6/r/n.计算帥-①亍的结果为（）/r/nA./r/n丄/r/nB./r/nC.-x/r/n2/r/ny/r/nD./r/n一小/r/nV/r/n7./r/n如图/r/n、/r/nABHCD/r/n、/r/nDE/r/n丄/r/nBE,BF/r/n、/r/nDF/r/n分别为/r/nZABE.ACDE/r/n的角平分线，/r/n()/r/n如图，把三角形纸片/r/nABC/r/n沿/r/nDE/r/n折叠，当点/r/nA/r/n落在四边形/r/nBCDE/r/n外部时，则/r/nZA/r/n与/r/nC.3ZA=2Z1-Z2/r/nZl.Z2/r/n之间的数量关系是(/r/nC.3ZA=2Z1-Z2/r/nB.3ZA=2(Z1-Z2)/r/nD/r/n・/r/nZA=Z1-Z2/r/n己知/r/nx/r/n2/r/n-4x-l=0＞/r/n则代数式/r/n2x(x—3)—(x—/r/n1/r/n尸+/r/n3/r/n的值为()/r/nTOC\o"1-5"\h\z/r/nA./r/n3/r/nB./r/n2/r/nC./r/n1/r/nD.-1/r/n已知/r/nx+y=5,xv=6,/r/n则的值是/r/n(/r/n)/r/nA.1B/r/n・/r/n13C/r/n・/r/n17D/r/n・/r/n25/r/n1/r/n2/r/n3/r/n11/r/n-/r/n式子：乔/r/n？/r/n正'莎的最简公分母是/r/n(/r/n)/r/nD.a/r/n8/r/n-b/r/ns/r/nA./r/n24x/r/n2/r/ny/r/n2/r/nxy/r/nB./r/n24/r/nx/r/n2/r/ny/r/n2/r/nC.12F/r/nD.a/r/n8/r/n-b/r/ns/r/n12./r/n计算：/r/n(a/r/n—/r/nb)(a/r/n+/r/nb)(a/r/n2/r/n+/r/nb/r/n2/r/n)(a/r/n4/r/n—/r/nb/r/n4/r/n)/r/n的结果是(/r/n)/r/nA.a/r/ns/r/n+2a/r/n4/r/nb/r/n4/r/n+b/r/ns/r/nB.a/r/ns/r/n-2a/r/n4/r/nb/r/n4/r/n+b/r/ns/r/nC.a/r/ns/r/n+b/r/ns/r/n二填空题/r/n13/r/n・/r/n已知//r/nb,c/r/n是/r/nZkABC/r/n的三边长，/r/na,b/r/n满足/r/n|a-7|+(b-1)/r/n2/r/n=0,c/r/n为奇数，则/r/nc=/r/n•/r/n14./r/n如图所示，过正五边形/r/nA3CDE/r/n的顶点/r/n3/r/n作一条射线与其内角/r/nZE43/r/n的角平分线相交/r/n于点/r/nP/r/n，/r/n且/r/nZABP=60°,/r/n则/r/n度./r/n15/r/n・/r/n已知射线/r/nOM/r/n•以/r/nO/r/n为圆心，任意长为半径画弧，与射线/r/nOM/r/n交于点/r/nA,/r/n再以点/r/nA/r/n为圆/r/n两弧交于点/r/nE,/r/n画射线/r/nOE,/r/n如图所示,/r/n则/r/n两弧交于点/r/nE,/r/n画射线/r/nOE,/r/n如图所示,/r/n则/r/nZAOB=/r/n(度)/r/nY—/r/n2/r/n16/r/n-/r/n当时’分式后的值为零./r/n一个等腰三角形的两边长分别为/r/n4cm/r/n和/r/n9cm,/r/n则它的周长为/r/n_cm/r/n・/r/n若关于/r/nx/r/n的方程/r/n—/r/n—无解，则仆/r/nx_5/r/n10-2x/r/n关于/r/nX/r/n的分式方程上$=/r/n1/r/n的解为负数，则/r/nd/r/n的取值范围是/r/n•/r/nX+1/r/n如果一个正多边形每一个内角都等于/r/n144/r/n。，那么这个正多边形的边数是—/r/n./r/n三、解答题/r/n说明代数式/r/n[U-y)/r/n2/r/n-U+y)U-y)]-(-2/r/n)/r/n-)+/r/n的值，与/r/ny/r/n的值无关./r/n解分式方程：/r/n•/r/nx+2/r/nJT/r/n一/r/n4/r/n如图，在/r/nAABC/r/n和/r/nZkAED/r/n中，/r/nAC/r/n与/r/nED/r/n相交于点/r/nE,AD=BC,ZDAB=ZCBA,/r/n求/r/niiE/r/n：/r/nAC=BD/r/n・/r/n为了对学生进行革命传统教育，红旗中学开展了“清明节祭扫”活动.全校学生从学校同时出发，步行/r/n4000/r/n米到达烈士纪念馆.学校要求九⑴班提前到达目的地，做好活动的准备工作.行走过程中，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，结果比其他班提前/r/n10/r/n分钟到达.分别求九/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【参考答案】杯*试卷处理标记,请不要删除/r/n选择题/r/n1/r/n./r/nA/r/n解析：/r/nA/r/n【解析】/r/n分析：根据多边形的内角和公式计算即可./r/n详解：/r/n»V-/r/n1/r/nW)/r/n=360°/r/n-T/r/n40°/r/n=9./r/n答:这个正多边形的边数是/r/n9/r/n.故选/r/nA./r/n点睛：本题考查了多边形，熟练掌握多边形的内角和公式是解答本题的关键./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/n先根据/r/nEC/r/n的垂直平分线交/r/nBD/r/n于点/r/nE/r/n证明/r/naBFE^aCFE/r/n(SAS)/r/n,/r/n根据全等三角形的性质和角平分线的性质得到/r/nZA/r/n恥二/r/n=/r/n,再根据三角形内角和定理即可得到/r/n答案./r/n【详解】/r/n解：如图：/r/nVBC/r/n的垂直平分线交/r/nBD/r/n于点/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(全等三角形对应角相等)，/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(三角形内角和定理)，/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故选/r/nC./r/n【点睛】/r/n本题主要考查了三角形全等的判定与性质、角平分线的性质、三角形内角和定理，证明/r/nZABE/r/n=/r/nZEBF/r/n=/r/nZECF/r/n是解题的关键./r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n根据最简分式的定义：分子和分母中不含公分母的分式，叫做最简分式，对四个选项中的/r/n

/r/n分式一一判断即可得出答案./r/n【详解】/r/n/+卩/r/n2/r/n解：/r/nA/r/n——/r/n,分式的分子与分母不含公因式，是最简分式；/r/nx/r/n+/r/ny/r/n6h3h/r/nB.—=—,/r/n分式的分子与分母含公因式/r/n2,/r/n不是最简分式；/r/n4a/r/n2a/r/n尢/r/n2—4/r/nC/r/n+2,/r/n分式的分子与分母含公因式心/r/n2,/r/n不是最简分式；/r/nX-/r/n2/r/n(/r/n1^+/r/n4(/./r/nD/r/n=q+4,/r/n分式的分子与分母含公因式/r/n4/r/n不是最简分式，/r/na/r/n故选/r/nA./r/n【点睛】/r/n本题考查了最简分式的概念•对每个分式的分子和分母分别进行因式分解是解题的关键./r/n4/r/n./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/n分别表示出客车在普通公路和高速公路上行驶的时间，即可得到方程./r/n【详解】/r/n根据题意：客车在普通公路上行驶的时间是竺小时，在高速公路上行驶的时间是空/r/nx/r/n2x/r/n小时，由所需时间比走普通公路所需时间少/r/n5/r/n小时可列方程：竽/r/n+5=—,/r/n2x/r/nx/r/n故选：/r/nC./r/n【点睛】/r/n此题考查分式方程的实际应用，正确理解题意找到等量关系是解题的关键./r/nD/r/n解析：/r/nD/r/n【解析】/r/n分析：根据全等三角形的判定定理/r/nAAS,/r/n可知应选/r/nD./r/n详解：解：如图:/r/n详解：解：如图:/r/n4/r/n选项中根据/r/nAB=DE,BC=EF/r/n,/r/n4=ND/r/n不能判定两个三角形全等，故/r/nA/r/n错：/r/nB/r/n选项三个角相等，不能判定两个三角形全等，故/r/nE/r/n错；/r/nC/r/n选项看似可用“边角边”定理判定两三角形全等，而对照图形可发现它们并不符合此判定条件，故/r/nC/r/n错；/r/nD/r/n选项中根据/r/n“4AS”/r/n可判定两个三角形全等，故选/r/nD/r/n：/r/n点睛：本题考查了全等三角形的条件，本题没有给出图形，增加此题的难度.若能顺利画出图形，对照图形和选项即可得到正确选项./r/nC/r/n解析:/r/nC/r/n【解析】/r/n【分析】/r/n根据分式的减法和除法可以解答本题/r/n【详解】/r/n=-x(x-y)-/r/nx-y/r/n=-x/r/n2/r/ny/r/n故答案为/r/nc/r/n【点睛】/r/n本题考查分式的混合运算，解答本题的关键是明确分式混合运算的计算方法./r/nD/r/n解析：/r/nD/r/n【解析】/r/n【分析】/r/n【详解】/r/n如图所示，过£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分别为/r/nZABE,ZCDE/r/n的角平分线，/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故选/r/nD./r/n【点睛】/r/n本题主要考查了平行线的性质以及角平分线的定义的运用，解题时注意：两直线平行，同/r/n旁内角互补.解决问题的关键是作平行线./r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n根据折叠的性质可得/r/nZAJZA,/r/n根据平角等于/r/n180/r/n。用/r/nZ1/r/n表示出/r/nZADA\/r/n根据三角形的一个外角等于与它不相邻的两个内角的和，用/r/nZ2/r/n与/r/nZA/r/n，/r/n表示出/r/nZ3,/r/n然后利用三角形的内角和等于/r/n180/r/n。列式整理即可得解./r/n【详解】/r/n如图所示:/r/n•/r/n・/r/n•AA^DE/r/n是/r/n△/r/nADE/r/n沿/r/nDE/r/n折叠得到，/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故选/r/nA./r/n【点睛】/r/n考查了三角形的内角和定理以及折叠的性质，根据折叠的性质，平角的定义以及三角形的一个外角等于与它不相邻的两个内角的和的性质，把/r/nZl/r/n、/r/nZ2/r/n、/r/nZA/r/n转化到同一个三角形中是解题的关键./r/n9./r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n先将原代数式进行去括号化简得出/r/nF/r/n-4/+2/r/n，/r/n然后根据/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最后代入计算即可./r/n【详解】/r/n由题意得：/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故选：/r/nA./r/n【点睛】/r/n本题主要考查了整式的化简求值，整体代入是解题关键./r/n10/r/n./r/nB/r/n解析：/r/nB/r/n【解析】/r/n【分析】/r/n将/r/nx+y=5/r/n两边平方，利用完全平方公式化简，把/r/nxy/r/n的值代入计算，即可求出所求式子的值./r/n【详解】/r/n解/r/n：/r/n将/r/nx+y=5/r/n两边平方得：/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将/r/nxv=6/r/n代入得：/r/nx/r/n2/r/n+12+y/r/n2/r/n=25,/r/n则/r/n故选：/r/nB./r/n【点睛】/r/n此题考查了完全平方公式，熟练掌握完全平方公式是解本题的关键./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/n分母都是单项式，根据最简公分母的求法：系数取最大系数，不同字母取最高次幕，将它们相乘即可求得./r/n【详解】/r/n1/r/n2/r/n3/r/n式子：的最简公分母是：/r/n12F/r/n）/r/nd/r/n2x~y3x~4xy~/r/n故选：/r/nC./r/n【点睛】/r/n本题考查最简公分母的定义与求法./r/nD/r/n解析：/r/nD/r/n【解析】/r/n试题分析：根据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故选/r/nD/r/n考点：平方差公式/r/n二、填空题/r/n7/r/n【解析】【分析】根据非负数的性质列式求出/r/nab/r/n的值再根据三角形的任意两边之和大于第三边两边之差小于第三边求出/r/nc/r/n的取值范围再根据/r/nc/r/n是奇数求出/r/nc/r/n的值【详解】/r/nTab/r/n满足/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根据非负数的性质列式求出/r/na/r/n、/r/nb/r/n的值，再根据三角形的任意两边之和大于第三边，两边之差小于第三边求出/r/nc/r/n的取值范围，再根据/r/nc/r/n是奇数求出/r/nc/r/n的值./r/n【详解】/r/nVa,b/r/n满足/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为奇数，/r/n••/r/n・/r/nc=7,/r/n故答案为/r/n7./r/n【点睛】/r/n本题考查非负数的性质：偶次方，解题的关键是明确题意，明确三角形三边的关系./r/n66/r/n［/r/n解析】【分析】首先根据正五边形的性质得到度然后根据角平分线的定义得到度再利用三角形内角和定理得到的度数【详解】解：/r/n•・/r/n•五边形为正五边形.••度/r/n•・/r/n•是的角平分线.••度/r/n•・•・•・/r/n故答案为：/r/n66/r/n【点睛】本题考查了多/r/n解析：/r/n66/r/n【解析】/r/n【分析】/r/n首先根据正五边形的性质得到/r/n=/r/n度，然后根据角平分线的定义得到/r/nZPAB/r/n=54/r/n度，再利用三角形内角和定理得到/r/nZAPB/r/n的度数./r/n【详解】/r/n解：/r/n•・/r/n•五边形/r/nA8CQE/r/n为正五边形，/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的角平分线，/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故答案为：/r/n66./r/n【点睛】/r/n本题考查了多边形内角与外角，题目中还用到了角平分线的定义及三角形内角和定理./r/n60/r/n【解析】【分析】首先连接/r/nAB/r/n由题意易证得/r/nAAOB/r/n是等边三角形根据等边三角形的性质可求得/r/nZAOB/r/n的度数【详解】连接/r/nAB/r/n根据题意得：/r/n0B/r/n二/r/n0A/r/n二/r/nAB/r/n・/r/n•/r/n・△/r/nAOB/r/n是等边三角形二/r/nZAOB/r/n二/r/n60°/r/n故答案为：/r/n解析：/r/n60/r/n【解析】/r/n【分析】/r/n首先连接/r/nAB,/r/n由题意易证得/r/nAAOB/r/n是等边三角形，根据等边三角形的性质，可求得/r/nZAOB/r/n的度数./r/n【详解】/r/n连接/r/nAB,/r/n根据题意得：/r/nOB=OA=AB,/r/n:./r/n/\AOB/r/n是等边三角形，/r/n/.ZA(95=60°./r/n【点睛】/r/n本题考查了等边三角形的判定与性质.此题难度不人，解题的关键是能根据题意得到/r/nOB=OA=AB./r/n16./r/n2/r/n【解析】由题意得：解得：/r/nx=2/r/n故答案为/r/n2/r/n解析：/r/n2/r/n【解析】/r/nx—/r/n2=0/r/n由题意得：/r/n｛cC/r/n，解得：/r/nx=2/r/n.故答案为/r/n2/r/nx+2h/r/n0/r/n17/r/n•/r/n22/r/n【解析】【分析】底边可能是/r/n4/r/n也可能是/r/n9/r/n分类讨论去掉不合条件的然后可求周长【详解】试题解析：①当腰是/r/n4cm/r/n底边是/r/n9cm/r/n时：不满足三角形的三边关系因此舍去②当底边是/r/n4cm/r/n腰长是/r/n9cm/r/n时能构成三角形则/r/n解析：/r/n22/r/n【解析】/r/n【分析】/r/n底边可能是/r/n4,/r/n也可能是/r/n9,/r/n分类讨论，去掉不合条件的，然后可求周长./r/n【详解】/r/n试题解析：①当腰是/r/n4cm,/r/n底边是/r/n9cm/r/n时：不满足三角形的三边关系，因此舍去./r/n②当底边是/r/n4cm,/r/n腰长是/r/n9cm/r/n时，能构成三角形，则其周长/r/n=4+9+9=22cm./r/n故填/r/n22./r/n【点睛】/r/n本题考查了等腰三角形的性质和三角形的三边关系：已知没有明确腰和底边的题目一定要想到两种情况，分类进行讨论，还应验证各种情况是否能构成三角形进行解答./r/n18/r/n・/r/n-8/r/n【解析】【分析】试题分析：・・•关于/r/nx/r/n的方程无解/r/n・・・/r/nx/r/n二/r/n5/r/n将分式方程去/r/n分母得/r/n:/r/n将/r/nx/r/n二/r/n5/r/n代入得：/r/nm=-8［/r/n详解】请在此输入详解！/r/n解析：/r/n・/r/n8/r/n【解析】/r/n【分析】/r/n试题分析：/r/n••/r/n•关于/r/nx/r/n的方程口=」一无解,/r/nAx=5/r/nx-5/r/n10-2x/r/n将分式方程口/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将/r/nx=5/r/n代入得：/r/nm=/r/n-8/r/n【详解】/r/n请在此输入详解！/r/n19./r/n【解析】【分析】分式方程去分母转化为整式方程由分式方程的解为负数求出/r/na/r/n的范围即可【详解】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由分式方程解为负数得到/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分式方程去分母转化为整式方程，由分式方程的解为负数，求出/r/na/r/n的范/r/nI/r/n制即可/r/n【详解】/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由分式方程解为负数，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故答案为:/r/na>l/r/n且/r/naH2/r/n【点睛】/r/n此题考查分式方程的解，解题关键在于求出/r/nx/r/n的值再进行分析/r/n20/r/n・/r/n10/r/n【解析】【分析】设正多边形的边数为/r/nn/r/n然后根据多边形的内角和公式列方程求解即可【详解】解：设正多边形的边数为/r/nn/r/n由题意得二/r/n144°/r/n解得/r/n210/r/n故答案为/r/n10/r/n【点睛】本题考查了多边形的内角与外角熟记公式/r/n解析：/r/n10/r/n【解析】/r/n【分析】/r/n设正多边形的边数为/r/n11,/r/n然后根据多边形的内角和公式列方程求解即可./r/n【详解】/r/n解：设正多边形的边数为/r/nm/r/n由题意得，/r/n（〃_/r/n2）・180/r/n=]44/r/n。,/r/nn/r/n解得/r/nn=10./r/n故答案为/r/n10./r/n【点睛】/r/n本题考查了多边形的内角与外角，熟记公式并准确列出方程是解题的关键./r/n三、解答题/r/n说明见解析./r/n【解析】/r/n试题分析：根据整式的混合运算的法则和顺序，先算完全平方和平方差，然后合并同类项化简，通过关化简可判断./r/n试题解析：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代数式的值与/r/ny/r/n无关./r/n原方程无解/r/n【解析】/r/n【分析】/r/n先找出方程的最简公分母，然后方程两边的每一项去乘最简公分母，化为整式方程，再求解，注意分式方程要检验./r/n【详解】/r/n方程两边同乘以/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检验：当/r/nx=-2/r/n时，/r/n(/r/nx+2)(x-2)/r/n=0,/r/n所以/r/