呼和浩特专版2020中考数学复习方案第一单元数与式课时训练03分式试题.docx

课时训练(三)分式(限时:30分钟)|夯实基础|1.2019扬州分式13-x可变形为()A.13+xB.-13+xC.1x-3D.-1x-32.2019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03.2019江西计算1a-1a2的结果为()A.aB.-aC.-1a3D.1a34.如果把分式x+2yx+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2倍B.缩小到原来的12C.是原来的23D.不变5.2019兰州化简:a2+1a+1-2a+1=()A.a-1B.a+1C.a-1a+1D.1a+16.一项工程,甲单独干,完成需要a天,乙单独干,完成需要b天,若甲、乙合作,完成这项工程所需的天数是()A.aba+bB.1a+1C.a+babD.ab(a+b)7.如果x2=y3=z40,那么x+y+zx+y-z的值是()A.7B.8C.9D.108.2018南充已知1x-1y=3,则代数式2x+3xy-2yx-xy-y的值是()A.-72B.-112C.92D.349.2019河北如图K3-1,若x为正整数,则表示(x+2)2x2+4x+4-1x+1的值的点落在()图K3-1A.段B.段C.段D.段10.2019泰州若分式12x-1有意义,则x的取值范围是.11.2019山西化简2xx-1-x1-x的结果是.12.2019梧州化简:2a2-8a+2-a=.13.若a,b互为倒数,则代数式a2+2ab+b2a+b1a+1b的值为.14.计算:(1)1-2x-1x-3x2-1;(2)a2-b2ab-ab-b2ab-a2.15.若x2=y3=z-50,求2x+y+3z2x的值.16.2019常德先化简,再选一个合适的数代入求值:x-1x2+x-x-3x2-12x2+x+1x2-x-1.17.2019苏州先化简,再求值:x-3x2+6x+91-6x+3,其中x=2-3.18.2019本溪先化简,再求值:a2-4a2-4a+4-12-a2a2-2a.其中a满足a2+3a-2=0.|拓展提升|19.2019孝感已知二元一次方程组x+y=1,2x+4y=9,则x2-2xy+y2x2-y2的值是()A.-5B.5C.-6D.620.2019达州a是不为1的有理数,我们把11-a称为a的差倒数,如2的差倒数为11-2=-1,-1的差倒数为11-(-1)=12,已知a1=5,a2是a1的差倒数,a3是a2的差倒数,a4是a3的差倒数,以此类推,a2019的值是()A.5B.-14C.43D.4521.2019滨州先化简,再求值:x2x-1-x2x2-1x2-xx2-2x+1,其中x是不等式组x-3(x-2)4,2x-335-x2的整数解.22.2019赤峰先化简,再求值:a2-2a+1a2-4a-1a-2+1a+2,其中a=|-1-3|-tan60+12-1.【参考答案】1.D2.B3.B4.D5.A解析a2+1a+1-2a+1=a2-1a+1=a-1.6.A解析 工作时间=工作总量工作效率.甲、乙一天的工作效率分别为1a,1b,则合作的工作效率为1a+1b,合作所需天数为11a+1b=aba+b.7.C解析 x2=y3=z40,可设x2=y3=z4=m(m0),则x=2m,y=3m,z=4m,x+y+zx+y-z=2m+3m+4m2m+3m-4m=9mm=9.8.D解析 1x-1y=3,y-x=3xy,x-y=-3xy,原式=2(x-y)+3xy(x-y)-xy=-6xy+3xy-3xy-xy=-3xy-4xy=34,故选D.9.B解析 (x+2)2x2+4x+4-1x+1=(x+2)2(x+2)2-1x+1=1-1x+1,根据x为正整数,类比反比例函数y=-k2+1x的性质,可得-12-1x+10,121-1x+11,表示(x+2)2x2+4x+4-1x+1的值的点落在段.10.x1211.3xx-112.a-4解析原式=2(a2-4)a+2-a=2(a+2)(a-2)a+2-a=2a-4-a=a-4.13.1解析a2+2ab+b2a+b1a+1b=(a+b)aba+b=ab,a,b互为倒数,ab=1,原式=1.14.解:(1)原式=x+1.(2)原式=ab.15.解:设x2=y3=z-5=k(k0),则x=2k,y=3k,z=-5k,所以2x+y+3z2x=22k+3k+3(-5k)22k=-8k4k=-2.16.解:原式=(x-1)2x(x+1)(x-1)-x(x-3)x(x+1)(x-1)2x2+x+1-x2+xx2-x=x+1x(x+1)(x-1)x(x-1)(x+1)2=1(x+1)2.取x=3代入1(x+1)2中,得原式=1(3+1)2=116.17.解: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=22.18.解:a2-4a2-4a+4-12-a2a2-2a=(a-2)(a+2)(a-2)2+1a-2a(a-2)2=a+2a-2+1a-2a(a-2)2=a+3a-2a(a-2)2=a(a+3)2=a2+3a2,a2+3a-2=0,a2+3a=2,原式=22=1.19.C解析解方程组得x=-52,y=72,所以x-y=-6,所以原式=(x-y)2(x+y)(x-y)=x-yx+y=-6,因此本题选C.20.D解析a1=5,a2是a1的差倒数,a2=11-5=-14,a3是a2的差倒数,a4是a3的差倒数,a3=11-(-14)=45,a4=11-45=5,根据规律可得an以5,-14,45为周期进行循环,2019=6733,a2019=45.21.解:原式=x3+x2(x+1)(x-1)-x2(x+1)(x-1)(x-1)2x(x-1)=x3(x+1)(x-1)(x-1)2x(x-1)=x2x+1,解不等式组,得1x3,则不等式组的整数解为1,2.当x=1时,原式无意义;当x=2时,原式=43.22.解:a2-2a+1a2-4a-1a-2+1a+2=(a-1)2(a+2)(a-2)a-2a-1+1a+2=a-1a+2+1a+2=aa+2,当a=|-1-3|-tan60+12-1=3+1-3+2=3时,原式=33+2=35.6