江西专版2020中考数学复习方案第一单元数与式课时训练04分式.docx

课时训练(四)分式(限时:40分钟)|夯实基础|1.2019宁波若分式1x-2有意义,则x的取值范围是()A.x2B.x2C.x0D.x-22.2019兰州化简:a2+1a+1-2a+1=()A.a-1B.a+1C.a-1a+1D.1a+13.2018葫芦岛若分式x2-1x+1的值为0,则x的值为()A.0B.1C.-1D.14.如图K4-1,下面的计算过程中,从哪一步开始出现错误()图K4-1A.B.C.D.5.2019河北如图K4-2,若x为正整数,则表示(x+2)2x2+4x+4-1x+1的值的点落在()图K4-2A.段B.段C.段D.段6.2019北京如果m+n=1,那么代数式2m+nm2-mn+1m(m2-n2)的值为()A.-3B.-1C.1D.37.2018南充已知1x-1y=3,则代数式2x+3xy-2yx-xy-y的值是()A.-72B.-112C.92D.348.2019北京若分式x-1x的值为0,则x的值为.9.2019吉林计算y2x2xy=.10.2019武汉计算2aa2-16-1a-4的结果是.11.化简:2a2-8a+2-a=.12.2019青岛化简:m-nmm2+n2m-2n.13.2019聊城化简:1-1a+3+6a2-9a+3a2-6a+9.14.2019广东先化简,再求值:xx-2-1x-2x2-xx2-4,其中x=2.15.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.16.先化简,再求值:a2-2ab+b2a2-b2a2-aba-2a+b,其中a,b满足(a-2)2+b+1=0.17.2019高安一模先化简1-1x-1x2-4x+4x2-1,再从不等式2x-16的正整数解中选一个适当的数代入求值.18.2019江西样卷四已知a2+4a=0,先化简,再求值:3a2-9+1a+3a2a-3.19.先化简:2x2+2xx2-1-x2-xx2-2x+1xx+1,然后解答下列问题:(1)当x=3时,求代数式的值.(2)原代数式的值能等于-1吗?为什么?|拓展提升|20.2019滨州观察下列一组数:a1=13,a2=35,a3=69,a4=1017,a5=1533,它们是按一定规律排列的,请利用其中规律,写出第n个数an=.(用含n的式子表示)【参考答案】1.B2.A3.B解析若分式的值为0,则x2-1=0,且x+10,x=1.故选B.4.B解析xx-y-yx+y=x(x+y)(x-y)(x+y)-y(x-y)(x-y)(x+y)=x2+xy-xy+y2(x-y)(x+y)=x2+y2x2-y2.故从第步开始出现错误.故选B.5.B解析(x+2)2x2+4x+4-1x+1=(x+2)2(x+2)2-1x+1=1-1x+1,x为正整数,-12-1x+10,121-1x+11,表示(x+2)2x2+4x+4-1x+1的值的点落在段.6.D解析2m+nm2-mn+1m(m2-n2)=2m+nm(m-n)+m-nm(m-n)(m+n)(m-n)=3mm(m-n)(m+n)(m-n)=3(m+n).m+n=1,原式=3.故选D.7.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.8.19.12x10.1a+411.a-412.解:原式=m-nmm(m-n)2=1m-n.13.解:原式=1-a-3+6(a+3)(a-3)(a-3)2a+3=1-a-3a+3=6a+3.14.解:原式=x-1x-2(x+2)(x-2)x(x-1)=x+2x.当x=2时,原式=2+22=2+1.15.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=1.16.解: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=-1.17.解:1-1x-1x2-4x+4x2-1=x-2x-1(x+1)(x-1)(x-2)2=x+1x-2.2x-16,2x7,x72,不等式2x-16的正整数解为1,2,3,又x1且x2,选x=3代入上式得:原式=3+13-2=4.18.解:3a2-9+1a+3a2a-3=a(a+3)(a-3)a2a-3=a(a+3)(a-3)a-3a2=1a2+3a.由a2+4a=0,可知a1=0,a2=-4.当a=0时,分式无意义,应舍去.当a=-4时,原式=116-12=14.19.解:原式=2x(x+1)(x+1)(x-1)-x(x-1)(x-1)2x+1x=2xx-1-xx-1x+1x=x+1x-1.(1)当x=3时,原式=2.(2)如果x+1x-1=-1,那么x+1=-x+1,x=0.当x=0时,除式xx+1=0,原代数式的值不能等于-1.20.n(n+1)2(2n+1)解析这组分数的分子分别为1,3=2+1,6=3+2+1,10=4+3+2+1,15=5+4+3+2+1,则第n个数的分子为n(n+1)2;分母分别为3=2+1,5=22+1,9=23+1,17=24+1,33=25+1,则第n个数的分母是2n+1,所以第n个数an=n(n+1)212n+1=n(n+1)2(2n+1).6