內(nèi)蒙古包頭市2019年中考數(shù)學(xué)總復(fù)習(xí) 第一單元 數(shù)與式 課時(shí)訓(xùn)練03 分式練習(xí)

課時(shí)訓(xùn)練(三) 分式A組·夯實(shí)基礎(chǔ)1.2018·白銀 若分式x2-4x的值為0,則x的值是()A.2或-2B.2C.-2D.02.2015·包頭樣題一 如果把分式2xx+y中的x,y都擴(kuò)大到原來的2倍,則分式的值()A.擴(kuò)大到原來的4倍B.擴(kuò)大到原來的2倍C.不變D.縮小到原來的123.2016·濱州 下列分式中,最簡分式是()A.x2-1x2+1B.x+1x2-1C.x2-2xy+y2x2-xyD.x2-362x+124.2018·威海 化簡(a-1)÷(1a-1)·a的結(jié)果是()A.-a2B.1C.a2D.-15.2018·淄博 化簡a2a-1-1-2a1-a的結(jié)果為()A.a+1a-1B.a-1C.aD.16.2018·蘇州 計(jì)算(1+1x)÷x2+2x+1x的結(jié)果是()A.x+1B.1x+1C.xx+1D.x+1x7.2018·內(nèi)江 已知1a-1b=13,則abb-a的值是()A.13B.-13C.3D.-38.2018·衡陽 計(jì)算:x2x+1-1x+1=. 9.2018·包頭樣題二 化簡:( m2m-1+11-m)·1m+1=. 10.2018·包頭樣題三 化簡a2+aba-b÷aba-b的結(jié)果是. 11.2015·包頭 化簡:a-2a-1a÷a2-1a=. 12.2017·東河區(qū)二模 化簡:a-32a-4÷(5a-2-a-2)=. 13.2018·瀘州 化簡:(1+2a-1)÷a2+2a+1a-1.14.2018·白銀 計(jì)算:ba2-b2÷(aa-b-1).15.2018·重慶B卷 計(jì)算:( a-1-4a-1a+1)÷a2-8a+16a+1.16.先化簡1x-2-2x·x2-2x2,再從0,1,2中選取一個合適的x的值代入求值.17.2018·濱州 先化簡,再求值:(xy2+x2y)·xx2+2xy+y2÷x2yx2-y2,其中x=0-(12)-1,y=2sin45°-8.B組·拓展提升18.2017·樂山 若a2-ab=0(b0),則aa+b=()A.0B.12C.0或12D.1或219.2018·南充 已知1x-1y=3,則代數(shù)式2x+3xy-2yx-xy-y的值是()A.-72B.-112C.92D.3420.已知x+1x=3,則下列三個等式:x2+1x2=7,x-1x=5,2x2-6x=-2,其中正確的個數(shù)為()A.0B.1C.2D.321.化簡aa2-4·a+2a2-3a-12-a,并求值,其中a與2,3構(gòu)成ABC的三邊,且a為整數(shù).22.先化簡,再求值:m-33m2-6m÷(m+2-5m-2),其中m是方程x2+2x-3=0的根.23.2017·鄂州 先化簡,再求值:( x-1+3-3xx+1)÷x2-xx+1,其中x的值從不等式組2-x3,2x-4<1的整數(shù)解中選取.參考答案1.A2.C3.A4.A5.B6.B7.C解析 1a-1b=b-aab=13,abb-a=3.故選擇C.8.x-19.110.a+bb11.a-1a+112.-12(a+3)13.解:原式=a-1+2a-1·a-1(a+1)2=1a+1.14.解:原式=b(a+b)(a-b)÷aa-b-a-ba-b=b(a+b)(a-b)÷a-a+ba-b=b(a+b)(a-b)÷ba-b=b(a+b)(a-b)·a-bb=1a+b.15.解:原式=a2-1-4a+1a+1·a+1(a-4)2=a2-4aa+1·a+1(a-4)2=aa-4.16.解:1x-2-2x·x2-2x2=xx(x-2)-2(x-2)x(x-2)·x(x-2)2=x-2(x-2)x(x-2)·x(x-2)2=x-2x+42=-x+42.由于x0且x2,因此只能取x=1.當(dāng)x=1時(shí),原式=-1+42=32.17.解:(xy2+x2y)·xx2+2xy+y2÷x2yx2-y2=xy(x+y)·x(x+y)2·(x+y)(x-y)x2y=x-y.當(dāng)x=0-(12)-1=1-2=-1,y=2sin45°-8=2×22-22=-2時(shí),原式=-1-(-2)=2-1.18.C解析 a2-ab=0(b0),a(a-b)=0,a=0或a-b=0,即a=0或a=b,aa+b=0或aa+b=12.19.D20.C解析 x+1x=3,x2+1x2=x+1x2-2=9-2=7,對;x-1x2=x+1x2-4=9-4=5,x-1x=±5,錯;2x2-6x=-2,2x2+2=6x.又x0,兩邊同時(shí)除以2x可得x+1x=3,對.21.解:原式=a(a+2)(a-2)·a+2a(a-3)+1a-2=1(a-2)(a-3)+a-3(a-2)(a-3)=a-2(a-2)(a-3)=1a-3.a與2,3構(gòu)成ABC的三邊,3-2<a<3+2,即1<a<5.a為整數(shù),a=2,3,4.當(dāng)a=2時(shí),分母2-a=0,舍去;當(dāng)a=3時(shí),分母a2-3a=0,舍去,故a的值只能為4.當(dāng)a=4時(shí),原式=14-3=1.22.解:原式=m-33m(m-2)÷m2-4m-2-5m-2=m-33m(m-2)·m-2(m+3)(m-3)=13m(m+3).m是方程x2+2x-3=0的根,m=-3或m=1.當(dāng)m=-3時(shí),原式無意義;當(dāng)m=1時(shí),原式=13m(m+3)=13×1×(1+3)=112.23.解析 先進(jìn)行分式的混合運(yùn)算,求出最簡結(jié)果;再解不等式組,從解集中確定出整數(shù)解,最后在整數(shù)解中選取一個使各個分式都有意義的x的值代入求值.解:原式=x2-1+3-3xx+1·x+1x(x-1)=(x-1)(x-2)x+1·x+1x(x-1)=x-2x.解不等式2-x3,得x-1;解不等式2x-4<1,得x<52,不等式組的解集為-1x<52,它的整數(shù)解為-1,0,1,2.x-1,0,1,x=2.當(dāng)x=2時(shí),原式=2-22=0.10