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2022年中考数学模拟试题汇编专题分式与分式方程 .pdf

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1、分式与分式方程一、选择题1(2016 浙江杭州萧山区模拟)下列等式成立的是（）AB（x1）（1x）=1x2CD（x1）2=x2+2x+1【考点】分式的混合运算；整式的混合运算【分析】利用分式的性质以及整式混合运算的计算方法逐一计算结果，进一步判断得出答案即可【解答】解：A、不能约分，此选项错误；B、（x1）（1x）=1+x2，此选项错误；C、=，此选项错误；D、（x1）2=x2+2x+1，此选项正确故选：D【点评】此题考查分式的混合运算，整式的混合运算，掌握分式的性质和整式混合运算的方法是解决问题的关键2、（2016 齐河三模）函数y=中自变量x 的取值范围是（）A、x 0 B、x 2 C、

2、x 3 D、x0，x 2 且 x3答案：D 3、（2016 齐河三模）若分式方程)2)(1(11xxmxx有增根，则m 的值为（）A、0 和 3 B、1 C、1 和-2 D、3 答案：D 4、（2016 齐河三模）解分式方程：+=1答案：1）去分母得：2+x（x+2）=x2 4，解得：x=3，检验：当x=3 时，（x+2）（x2）0，故 x=3 是原方程的根5、（2016 天津南开区二模）某工厂现在平均每天比原计划多生产50 台机器，现在生产600台所需时间与原计划生产450 台机器所需时间相同.设原计划平均每天生产x 台机器，根据题意，下面所列方程正确的是（）A=B=C=D=考点：分式方程

3、的应用答案：B试题解析：设原计划平均每天生产x 台机器，则实际平均每天生产（x+50）台机器，由题意得，=故选 B6、（2016 天津市南开区一模）化简的结果（）Ax1 Bx CD【考点】分式的乘除法【专题】计算题；分式【分析】原式利用除法法则变形，约分即可得到结果【解答】解：原式=?=x 1，故选 A【点评】此题考查了分式的乘除法，熟练掌握运算法则是解本题的关键7、(2016 重庆铜梁巴川一模）函数y=+中自变量 x 的取值范围是（）Ax2 Bx2 且 x1Cx2 且 x1Dx1【分析】根据二次根式的性质和分式的意义，被开方数大于等于0，分母不等于0，就可以求解【解答】解：根据二次根式有

4、意义，分式有意义得：2x0 且 x10，解得：x2 且 x1 故选：B8、(2016 重庆巴南一模）分式方程=0 的解为（）文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2

5、R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7

6、F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I1

7、0M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编

8、码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F

9、10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2

10、B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9文档编码:CR2F10D8A2B5 HM2D3H8Z2R3 ZD7F4M5I10M9Ax=3 Bx=5 Cx=5 D无解【分析

11、】分式方程去分母转化为整式方程，求出整式方程的解得到x 的值，经检验即可得到分式方程的解【解答】解：去分母得：2x+23x+3=0，解得：x=5，经检验 x=5 是分式方程的解故选 C 9、(2016 山西大同一模）在解分式方程1x+=2x-1x-1时，我们第一步通常是去分母，即方程两边同乘以最简公分母（x-1），把分式方程变形为整式方程求解。解决这个问题的方法用到的数学思想是（）A数形结合B转化思想C模型思想D特殊到一般答案：B 10、(2016 云南省曲靖市罗平县二模)方程1=的解集是（）A 3 B3 C4 D 4【考点】解分式方程【专题】计算题；分式方程及应用【分析】分式方程去分母

12、转化为整式方程，求出整式方程的解得到x 的值，经检验即可得到分式方程的解【解答】解：去分母得：3xx+4=1，解得：x=3，经检验 x=3 是分式方程的解故选 B【点评】此题考查了解分式方程，解分式方程的基本思想是“转化思想”，把分式方程转化为整式方程求解解分式方程一定注意要验根11、(2016 郑州二模)郑徐客运专线（简称郑徐高铁），即郑州至徐州高速铁路，是国家中长期铁路网规划中“四纵四横”之一的徐兰客运专线的重要组成部分，20XX 年 7 月将要开通运营高铁列车从郑州到徐州的运行时间比原普通车组的运行时间要快约1.4 个小时，已知郑州到徐州的铁路长约为361 千米，原普通车组列车的平均速

13、度为x 千米时，高铁列车的平均速度比原普通车组列车增加了145 千米时依题意，下面所列方程正确的是文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码

14、:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2

15、HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 Z

16、I2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档

17、编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V

18、2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7

19、 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10A4.1361145361xxB4.1145361361xxC4.1145361361xxD361)145(4.1xx答案：C 12.（2016江苏省南京市钟爱中学

20、九年级下学期期初考试）已知，C是线段 AB的黄金分割点，AC BC，若 AB=2，则 BC=（）A 1 B（+1）C 3 D（1）答案：A 13.（2016上海市闸北区中考数学质量检测4 月卷）下列代数式中，属于分式的是（）（A）3；（B）12ab；（C）1x；（D）34a b答案：C14（2016湖南省岳阳市十二校联考一模）式子有意义的x 的取值范围是（）Ax 且 x 1 Bx 1 CD【考点】二次根式有意义的条件；分式有意义的条件【分析】根据被开方数大于等于0，分母不等于0 列式进行计算即可得解【解答】解：根据题意得，2x+1 0 且 x1 0，解得 x 且 x 1故选 A【点评】本题考

21、查的知识点为：分式有意义，分母不为0；二次根式的被开方数是非负数15.（2016黑龙江大庆一模）化简xxx1112的结果是()A.1x-B.11x+C.1x+D.1xx-答案：C 16.（2016黑龙江齐齐哈尔一模）若关于x的分式方程1xmmx无解，则m的值为()A.0 B.2 C.0 或 2 D.2答案：C 文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P

22、2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8

23、C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:C

24、T5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR

25、3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2

26、G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码

27、:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2

28、HR3P2Q6Y4O7 ZI2G8C8R1H1017.（2016广东东莞联考）化简的结果是（）Ax+1 B x1 C x Dx【考点】分式的加减法【专题】计算题【分析】将分母化为同分母，通分，再将分子因式分解，约分【解答】解：=x，故选：D【点评】本题考查了分式的加减运算分式的加减运算中，如果是同分母分式，那么分母不变，把分子直接相加减即可；如果是异分母分式，则必须先通分，把异分母分式化为同分母分式，然后再相加减18.（2016广东深圳一模）A，B两地相距48 千米，一艘轮船从A地顺流航行至B地，又立即从B地逆流返回A地，共用去9 小时，已知水流速度为4 千米/时，若设该轮船在静水中的速度为x

29、千米/时，则可列方程（）AB C+4=9 D【考点】由实际问题抽象出分式方程【专题】应用题【分析】本题的等量关系为：顺流时间+逆流时间=9小时【解答】解：顺流时间为：；逆流时间为：所列方程为：+=9故选 A文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2

30、G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码

31、:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2

32、HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 Z

33、I2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档

34、编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V

35、2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10【点评】未知量是速度，有速度，一定是根据时间来列等量关系

36、的找到关键描述语，找到等量关系是解决问题的关键二、填空题1(2016四川峨眉二模）当x 时，分式12x有意义答案：2x2.（2016江苏丹阳市丹北片一模）3 的倒数是 ,16的平方根是 .答案：31，2；3.（2016江苏丹阳市丹北片一模）计算：38-=，分解因式：16x-x92=答案：-2，（3x-1）2；4.（2016江苏丹阳市丹北片一模）使x-2有意义的x 的取值范围是，使分式2x3-x的值为零的x 的值是答案：2x，x=3；5（2016辽宁丹东七中一模）函数y=21x中，自变量x 的取值范围是。答案：a2 6（2016 湖南省岳阳市十二校联考一模）分式方程=的解为x=9【考点】解分式方

37、程【专题】计算题【分析】分式方程去分母转化为整式方程，求出整式方程的解得到x 的值，经检验即可得到分式方程的解【解答】解：去分母得：4x=3x9，解得：x=9，经检验 x=9 是分式方程的解故答案为：x=9【点评】此题考查了解分式方程，解分式方程的基本思想是“转化思想”，把分式方程转化为整式方程求解解分式方程一定注意要验根文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V

38、2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7

39、 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10

40、文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B

41、9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4

42、O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H

43、10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H

44、2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H107（2016湖南湘潭一模）当x=_时，分式xx1的值为零答案：1 8.（2016河北石家庄一模）1=【考点】分式的加减法【专题】计算题；分式【分析】原式通分并利用同分母分式的减法法则计算即可得到结果【解答】解：原式=，故答案为：【点评】此题考查了分式的加减法，熟练掌握运算法则是解本题的关键9.（2016黑龙江大庆一模）已知实数m、n 满足mm222，nn222，则mnnm=_答案：-4 或 2 三、解答题1(2016 浙江镇江模拟)化简：11)1(aaaa.解：11)1(aaaa=1112aaaaa=)1(12aaa=2a2(201

45、6上海浦东模拟)（本题满分10 分）解方程：228224xxxxx解：去分母得：2228x xx整理得：220 xx解得：11x，22x经检验11x是原方程的根，22x是原方程的增根文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码

46、:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2

47、HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 Z

48、I2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档

49、编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V

50、2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7 ZI2G8C8R1H10文档编码:CT5L2H2B9V2 HR3P2Q6Y4O7

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关键词：: 2022年中考数学模拟试题汇编专题分式与分式方程 2022 年中数学模拟试题汇编专题分式方程

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