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课时训练(三) 分式A组夯实基础1.xx白银 若分式x2-4x的值为0,则x的值是()A.2或-2B.2C.-2D.02.xx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.xx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.xx威海 化简(a-1)(1a-1)a的结果是()A.-a2B.1C.a2D.-15.xx淄博 化简a2a-1-1-2a1-a的结果为()A.a+1a-1B.a-1C.aD.16.xx苏州 计算(1+1x)x2+2x+1x的结果是()A.x+1B.1x+1C.xx+1D.x+1x7.xx内江 已知1a-1b=13,则abb-a的值是()A.13B.-13C.3D.-38.xx衡阳 计算:x2x+1-1x+1=.9.xx包头样题二 化简:( m2m-1+11-m)1m+1=.10.xx包头样题三 化简a2+aba-baba-b的结果是.11.xx包头 化简:a-2a-1aa2-1a=.12.xx东河区二模 化简:a-32a-4(5a-2-a-2)=.13.xx泸州 化简:(1+2a-1)a2+2a+1a-1.14.xx白银 计算:ba2-b2(aa-b-1).15.xx重庆B卷 计算:( a-1-4a-1a+1)a2-8a+16a+1.16.先化简1x-2-2xx2-2x2,再从0,1,2中选取一个合适的x的值代入求值.17.xx滨州 先化简,再求值:(xy2+x2y)xx2+2xy+y2x2yx2-y2,其中x=0-(12)-1,y=2sin45-8.B组拓展提升18.xx乐山 若a2-ab=0(b0),则aa+b=()A.0B.12C.0或12D.1或219.xx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:x2+1x2=7,x-1x=5,2x2-6x=-2,其中正确的个数为()A.0B.1C.2D.321.化简aa2-4a+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.23.xx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-x3,2x-41的整数解中选取.参考答案1.A2.C3.A4.A5.B6.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-1a-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+1a+1(a-4)2=a2-4aa+1a+1(a-4)2=aa-4.16.解:1x-2-2xx2-2x2=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.当x=1时,原式=-1+42=32.17.解:(xy2+x2y)xx2+2xy+y2x2yx2-y2=xy(x+y)x(x+y)2(x+y)(x-y)x2y=x-y.当x=0-(12)-1=1-2=-1,y=2sin45-8=222-22=-2时,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,对;x-1x2=x+1x2-4=9-4=5,x-1x=5,错;2x2-6x=-2,2x2+2=6x.又x0,两边同时除以2x可得x+1x=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,3-2a3+2,即1a5.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)=131(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+1x+1x(x-1)=(x-1)(x-2)x+1x+1x(x-1)=x-2x.解不等式2-x3,得x-1;解不等式2x-41,得x52,不等式组的解集为-1x52,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.
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