中考数学总复习《分式》专项测试卷含答案

第第页中考数学总复习《分式》专项测试卷含答案学校:___________班级：___________姓名：___________考号：___________A层·基础过关1.(2024·崇左一模)若分式x2x+10有意义,则A.x≠0 B.x≠5C.x≠-5 D.x≠-102.计算3a+2aA.1a B.6a2 C.5a3.(2024·贵州)计算3aa−bA.3 B.3a+3bC.1 D.64.化简(1+1a−1)÷aA.a+1 B.a+1a C.a−1a5.(2024·山东)计算a2−4a÷(a+1-5A.a+2a−2 C.(a−2)(a+2)a6.若分式1x−2有意义,则实数x的取值范围是7.(2024·湖北)计算mm+1+1m8.已知分式|x|−32x−6=0,9.(2024·临夏州)化简:(a+1+1a−1)÷10.先化简,再求值:a−1a−2·a2−4a2−211.化简求值:x−1x2+2x+1÷1−2B层·能力提升12.已知x2-x-1=0,计算(2x+1-1x)÷A.1 B.-1 C.2 D.-213.(2024·眉山)已知a1=x+1(x≠0且x≠-1),a2=11−a1,a3=11−a2,…,an=11−an14.(2024·扬州)化简:x−2x+1÷(15.(2024·盐城)先化简,再求值:1-a−3a÷a2−9a16.(2024·江西)先化简,再求值:(1-1a+3)÷a+2a2C层·挑战冲A+17.化简(xx+1+xx−1)·(1)甲同学解法的依据是②,乙同学解法的依据是③;(填序号)

①等式的基本性质;②分式的基本性质;③乘法分配律;④乘法交换律.(2)请选择一种解法,写出完整的解答过程.参考答案A层·基础过关1.(2024·崇左一模)若分式x2x+10有意义,则A.x≠0 B.x≠5C.x≠-5 D.x≠-102.计算3a+2aA.1a B.6a2 C.5a3.(2024·贵州)计算3aa−bA.3 B.3a+3bC.1 D.64.化简(1+1a−1)÷aA.a+1 B.a+1a C.a−1a5.(2024·山东)计算a2−4a÷(a+1-5A.a+2a−2 C.(a−2)(a+2)a6.若分式1x−2有意义,则实数x的取值范围是x≠27.(2024·湖北)计算mm+1+1m+1的结果是8.已知分式|x|−32x−6=0,则x9.(2024·临夏州)化简:(a+1+1a−1)÷【解析】原式=(a+1)(=a2−1+1a−1·a=aa10.先化简,再求值:a−1a−2·a2−4a2−2【解析】原式=a−1a−2·=a+2a−1-2a当a=12时,原式=1211.化简求值:x−1x2+2x+1÷1−2【解析】原式=x−1(=x−1(x+1)2当x=3-1时,原式=13−1+1=B层·能力提升12.已知x2-x-1=0,计算(2x+1-1x)÷A.1 B.-1 C.2 D.-213.(2024·眉山)已知a1=x+1(x≠0且x≠-1),a2=11−a1,a3=11−a2,…,an=11−an−114.(2024·扬州)化简:x−2x+1÷(【解析】x−2x+1÷(x-2)=x−2x15.(2024·盐城)先化简,再求值:1-a−3a÷a2−9a【解析】原式=1-a−3a·a(a+1)(a+3)(当a=4时,原式=24+3=216.(2024·江西)先化简,再求值:(1-1a+3)÷a+2a2【解析】原式=a+2a+3÷a+2(a+3)(将a=1代入,得:原式=1-3=-2.C层·挑战冲A+17.化简(xx+1+xx−1)·(1)甲同学解法的依据是②,乙同学解法的依据是③;(填序号)

①等式的基本性质;②分式的基本性质;③乘法分配律;④乘法交换律.【解析】(1)甲同学的解法是:先把括号内两个分式通分后相加,再进行乘法运算通分的依据是分式的基本性质;乙同学的解法是:根据乘法的分配律,去掉括号后,先算分式的乘