中考數(shù)學(xué)總復(fù)習(xí)《分式》專項(xiàng)測(cè)試卷含答案

第第頁中考數(shù)學(xué)總復(fù)習(xí)《分式》專項(xiàng)測(cè)試卷含答案學(xué)校:___________班級(jí)：___________姓名：___________考號(hào)：___________A層·基礎(chǔ)過關(guān)1.(2024·崇左一模)若分式x2x+10有意義,則A.x≠0 B.x≠5C.x≠-5 D.x≠-102.計(jì)算3a+2aA.1a B.6a2 C.5a3.(2024·貴州)計(jì)算3aa?bA.3 B.3a+3bC.1 D.64.化簡(jiǎn)(1+1a?1)÷aA.a+1 B.a+1a C.a?1a5.(2024·山東)計(jì)算a2?4a÷(a+1-5A.a+2a?2 C.(a?2)(a+2)a6.若分式1x?2有意義,則實(shí)數(shù)x的取值范圍是7.(2024·湖北)計(jì)算mm+1+1m8.已知分式|x|?32x?6=0,9.(2024·臨夏州)化簡(jiǎn):(a+1+1a?1)÷10.先化簡(jiǎn),再求值:a?1a?2·a2?4a2?211.化簡(jiǎn)求值:x?1x2+2x+1÷1?2B層·能力提升12.已知x2-x-1=0,計(jì)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·揚(yáng)州)化簡(jiǎn):x?2x+1÷(15.(2024·鹽城)先化簡(jiǎn),再求值:1-a?3a÷a2?9a16.(2024·江西)先化簡(jiǎn),再求值:(1-1a+3)÷a+2a2C層·挑戰(zhàn)沖A+17.化簡(jiǎn)(xx+1+xx?1)·(1)甲同學(xué)解法的依據(jù)是②,乙同學(xué)解法的依據(jù)是③;(填序號(hào))

①等式的基本性質(zhì);②分式的基本性質(zhì);③乘法分配律;④乘法交換律.(2)請(qǐng)選擇一種解法,寫出完整的解答過程.參考答案A層·基礎(chǔ)過關(guān)1.(2024·崇左一模)若分式x2x+10有意義,則A.x≠0 B.x≠5C.x≠-5 D.x≠-102.計(jì)算3a+2aA.1a B.6a2 C.5a3.(2024·貴州)計(jì)算3aa?bA.3 B.3a+3bC.1 D.64.化簡(jiǎn)(1+1a?1)÷aA.a+1 B.a+1a C.a?1a5.(2024·山東)計(jì)算a2?4a÷(a+1-5A.a+2a?2 C.(a?2)(a+2)a6.若分式1x?2有意義,則實(shí)數(shù)x的取值范圍是x≠27.(2024·湖北)計(jì)算mm+1+1m+1的結(jié)果是8.已知分式|x|?32x?6=0,則x9.(2024·臨夏州)化簡(jiǎn):(a+1+1a?1)÷【解析】原式=(a+1)(=a2?1+1a?1·a=aa10.先化簡(jiǎn),再求值:a?1a?2·a2?4a2?2【解析】原式=a?1a?2·=a+2a?1-2a當(dāng)a=12時(shí),原式=1211.化簡(jiǎn)求值:x?1x2+2x+1÷1?2【解析】原式=x?1(=x?1(x+1)2當(dāng)x=3-1時(shí),原式=13?1+1=B層·能力提升12.已知x2-x-1=0,計(jì)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·揚(yáng)州)化簡(jiǎn):x?2x+1÷(【解析】x?2x+1÷(x-2)=x?2x15.(2024·鹽城)先化簡(jiǎn),再求值:1-a?3a÷a2?9a【解析】原式=1-a?3a·a(a+1)(a+3)(當(dāng)a=4時(shí),原式=24+3=216.(2024·江西)先化簡(jiǎn),再求值:(1-1a+3)÷a+2a2【解析】原式=a+2a+3÷a+2(a+3)(將a=1代入,得:原式=1-3=-2.C層·挑戰(zhàn)沖A+17.化簡(jiǎn)(xx+1+xx?1)·(1)甲同學(xué)解法的依據(jù)是②,乙同學(xué)解法的依據(jù)是③;(填序號(hào))

①等式的基本性質(zhì);②分式的基本性質(zhì);③乘法分配律;④乘法交換律.【解析】(1)甲同學(xué)的解法是:先把括號(hào)內(nèi)兩個(gè)分式通分后相加,再進(jìn)行乘法運(yùn)算通分的依據(jù)是分式的基本性質(zhì);乙同學(xué)的解法是:根據(jù)乘法的分配律,去掉括號(hào)后,先算分式的乘