内蒙古包头市2022年中考数学总复习第一单元数与式课时训练03分式练习

课时训练(三)　分式A组·夯实基础1.[2022·白银]若分式x2-4x的值为0,则x的值是(　　)A.2或-2B.2C.-2D.02.[2022·包头样题一]如果把分式2xx+y中的x,y都扩大到原来的2倍,则分式的值(　　)A.扩大到原来的4倍B.扩大到原来的2倍C.不变D.缩小到原来的123.[2022·滨州]下列分式中,最简分式是(　　)A.x2-1x2+1B.x+1x2-1C.x2-2xy+y2x2-xyD.x2-362x+124.[2022·威海]化简(a-1)÷(1a-1)·a的结果是(　　)A.-a2B.1C.a2D.-15.[2022·淄博]化简a2a-1-1-2a1-a的结果为(　　)A.a+1a-1B.a-1C.aD.16.[2022·苏州]计算(1+1x)÷x2+2x+1x的结果是(　　)A.x+1B.1x+110\nC.xx+1D.x+1x7.[2022·内江]已知1a-1b=13,则abb-a的值是(　　)A.13B.-13C.3D.-38.[2022·衡阳]计算:x2x+1-1x+1=　　　　. 9.[2022·包头样题二]化简:(m2m-1+11-m)·1m+1=　　　　. 10.[2022·包头样题三]化简a2+aba-b÷aba-b的结果是　　　　. 11.[2022·包头]化简:a-2a-1a÷a2-1a=　　　　. 12.[2022·东河区二模]化简:a-32a-4÷(5a-2-a-2)=　　　　. 13.[2022·泸州]化简:(1+2a-1)÷a2+2a+1a-1.14.[2022·白银]计算:ba2-b2÷(aa-b-1).10\n15.[2022·重庆B卷]计算:(a-1-4a-1a+1)÷a2-8a+16a+1.16.先化简1x-2-2x·x2-2x2,再从0,1,2中选取一个合适的x的值代入求值.10\n17.[2022·滨州]先化简,再求值:(xy2+x2y)·xx2+2xy+y2÷x2yx2-y2,其中x=π0-(12)-1,y=2sin45°-8.B组·拓展提升10\n18.[2022·乐山]若a2-ab=0(b≠0),则aa+b=(　　)A.0B.12C.0或12D.1或219.[2022·南充]已知1x-1y=3,则代数式2x+3xy-2yx-xy-y的值是(　　)A.-72B.-112C.92D.3420.已知x+1x=3,则下列三个等式:①x2+1x2=7,②x-1x=5,③2x2-6x=-2,其中正确的个数为(　　)A.0B.1C.2D.321.化简aa2-4·a+2a2-3a-12-a,并求值,其中a与2,3构成△ABC的三边,且a为整数.22.先化简,再求值:m-33m2-6m÷(m+2-5m-2),其中m是方程x2+2x-3=0的根.10\n23.[2022·鄂州]先化简,再求值:(x-1+3-3xx+1)÷x2-xx+1,其中x的值从不等式组2-x≤3,2x-4<1的整数解中选取.10\n10\n参考答案1.A　2.C　3.A　4.A　5.B　6.B7.C　[解析]∵1a-1b=b-aab=13,∴abb-a=3.故选择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)210\n=x-2x+42=-x+42.由于x≠0且x≠2,因此只能取x=1.当x=1时,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.当x=π0-(12)-1=1-2=-1,y=2sin45°-8=2×22-22=-2时,原式=-1-(-2)=2-1.18.C　[解析]∵a2-ab=0(b≠0),∴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=7,①对;∵x-1x2=x+1x2-4=9-4=5,∴x-1x=±5,②错;∵2x2-6x=-2,∴2x2+2=6x.又∵x≠0,∴两边同时除以2x可得x+1x=3,③对.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与2,3构成△ABC的三边,10\n∴3-2<a<3+2,即1<a<5.∵a为整数,∴a=2,3,4.当a=2时,分母2-a=0,舍去;当a=3时,分母a2-3a=0,舍去,故a的值只能为4.∴当a=4时,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.当m=-3时,原式无意义;当m=1时,原式=13m(m+3)=13×1×(1+3)=112.23.[解析]先进行分式的混合运算,求出最简结果;再解不等式组,从解集中确定出整数解,最后在整数解中选取一个使各个分式都有意义的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-x≤3,得x≥-1;解不等式2x-4<1,得x<52,∴不等式组的解集为-1≤x<52,它的整数解为-1,0,1,2.∵x≠-1,0,1,∴x=2.当x=2时,原式=2-22=0.10