包头专版2020年中考数学复习第一单元数与式课时训练04分式及其运算.docx

第4课时分式及其运算|夯实基础|1.如果式子x+1x-3有意义,则x的取值范围是()A.x-1B.x-1且x3C.x-1且x3D.x-1且x32.2019聊城如果分式|x|-1x+1的值为0,那么x的值为()A.-1B.1C.-1或1D.1或03.下列约分正确的是()A.x6x2=x3B.x-yx-y=0C.x-yx2-xy=1xD.2x2y4xy2=124.下列分式中,最简分式是()A.x2-1x2+1B.x+1x2-1C.x2-2xy+y2x2-xyD.x2-362x+125.2019陇南下面的计算过程中,从哪一步开始出现错误()图4-1A.B.C.D.6.2019兰州化简:a2+1a+1-2a+1=()A.a-1B.a+1C.a-1a+1D.1a+17.若y=-x+3,且xy,则x2x-y+y2y-x的值为()A.3B.-3C.13D.-138.2019眉山化简a-b2aa-ba的结果是()A.a-bB.a+bC.1a-bD.1a+b9.2016包头 化简1a+1b1a2-1b2ab,其结果是()A.a2b2a-bB.a2b2b-aC.1a-bD.1b-a10.2019临沂计算a2a-1-a-1的正确结果是()A.-1a-1B.1a-1C.-2a-1a-1D.2a-1a-111.2019青山区二模已知A1+1x-1=1x+1,则A=()A.x-1x2+xB.xx2-1C.1x2-1D.x2-112.已知a为整数,且a+1a-3-a-3a+2a2-6a+9a2-4为正整数,则所有符合条件的a的值的和为()A.0B.12C.10D.813.2019攀枝花一辆货车送货上山,并按原路下山.上山速度为a千米/时,下山速度为b千米/时,则货车上、下山的平均速度为()A.12(a+b)B.aba+bC.a+b2abD.2aba+b14.2019包头样题二计算1x+1-x2x+1的结果为.15.2019武汉计算2aa2-16-1a-4的结果是.16.2019包头样题三化简:x2-5x+2x+2+1x2-4x2+4x+4的值为.17.2019包头一模若x=2-1,则代数式1x+x2-2xx-1x2+x的值为.18.2019昆都仑区二模若x=tan45+12-1,则代数式1-x+3x+1x2+4x+4x+1的值为.19.2019杭州化简:4xx2-4-2x-2-1.圆圆的解答如下:4xx2-4-2x-2-1=4x-2(x+2)-(x2-4)=-x2+2x.圆圆的解答正确吗?如果不正确,写出正确的解答.20.2019泸州化简:m+2+1mmm+1.21.(1)2019长沙先化简,再求值:a+3a-1-1a-1a2+4a+4a2-a,其中a=3.(2)2019苏州先化简,再求值:x-3x2+6x+91-6x+3,其中x=2-3.(3)2019娄底先化简,再求值:a2-2ab+b2a-b+1b-1a,其中a=2-1,b=2+1.(4)2019遂宁先化简,再求值:a2-2ab+b2a2-b2a2-aba-2a+b,其中a,b满足(a-2)2+b+1=0.22.2019遵义化简式子a2-2aa2-4a+4+1a2-1a2+a,并在-2,-1,0,1,2中选取一个合适的数作为a的值代入求值.|拓展提升|23.已知2yx-xy=3,则3x2-2xy-6y22y2-x2=.24.(1)2019德州先化简,再求值:2m-1nm2+n2mn-5nmm2n+2nm+2,其中m+1+(n-3)2=0.(2)2019泰安先化简,再求值:a-9+25a+1a-1-4a-1a+1,其中a=2.【参考答案】1.C2.B3.C4.A5.B6.A7.A8.B9.B解析1a+1b1a2-1b2ab=a+babb2-a2a2b2ab=a+bab(b+a)(b-a)a2b2ab=a+baba2b2(b+a)(b-a)ab=a2b2b-a.故选B.10.B11.B12.C解析a+1a-3-a-3a+2a2-6a+9a2-4=a+1a-3-a-3a+2(a-3)2(a+2)(a-2)=a+1a-3-a-3a+2(a+2)(a-2)(a-3)2=a+1a-3-a-2a-3=3a-3,a为整数,且分式的值为正整数,a-3=1或a-3=3,a=4或a=6,所有符合条件的a的值的和为4+6=10.13.D解析设山路全程为1,则货车上山所用时间为1a,下山所用时间为1b.货车上、下山的平均速度=21a+1b=2aba+b,故选D.14.1-x15.1a+4解析原式=2a(a+4)(a-4)-a+4(a+4)(a-4)=2a-a-4(a+4)(a-4)=a-4(a+4)(a-4)=1a+4.16.x-217.218.-1519.解:圆圆的解答错误,正确解法:4xx2-4-2x-2-1=4x(x-2)(x+2)-2(x+2)(x-2)(x+2)-(x-2)(x+2)(x-2)(x+2)=4x-2x-4-x2+4(x-2)(x+2)=2x-x2(x-2)(x+2)=-xx+2.20.解:原式=m2+2m+1mmm+1=(m+1)2mmm+1=m+1.21.解:(1)原式=a+2a-1a(a-1)(a+2)2=aa+2,当a=3时,原式=33+2=35.(2)原式=x-3(x+3)2x-3x+3=x-3(x+3)2x+3x-3=1x+3,当x=2-3时,原式=12-3+3=12=22.(3)a=2-1,b=2+1,a-b=(2-1)-(2+1)=-2,ab=(2-1)(2+1)=1.原式=(a-b)2a-b+a-bab=a-b+a-bab=-2+-21=-4.(4)原式=(a-b)2(a+b)(a-b)a(a-b)a-2a+b=a-ba+b1a-b-2a+b=-1a+b.(a-2)2+b+1=0,a=2,b=-1,原式=-1.22.解:原式=a(a-2)(a-2)2+1(a+1)(a-1)a(a+1)=a+a-2a-2a-1a=2(a-1)a-2aa-1=2aa-2.a-1,0,1,2,a=-2,当a=-2时,原式=1.23.-113解析2yx-xy=3,2y2-x2xy=3,2y2-x2=3xy,原式=3x2-2xy-6y23xy=xy-23-2yx=-3-23=-113.24.解:(1)2m-1nm2+n2mn-5nmm2n+2nm+2=2n-mmnm2+n2-5n2mnm2+4n2+4mn2mn=2n-mmnmn(m+2n)(m-2n)(m+2n)22mn=-m+2n2mn.m+1+(n-3)2=0,m+1=0,n-3=0,m=-1,n=3,-m+2n2mn=-1+232(-1)3=56,原式的值为56.(2