安徽省2022年中考数学一轮复习第一讲数与代数第一章数与代数1.3分式测试

1.3　分式[过关演练]　(30分钟　　80分)1.(2022·甘肃白银)若分式x2-4x的值为0,则x的值是(A)A.2或-2B.2C.-2D.0【解析】∵分式x2-4x的值为0,∴x2-4=0,解得x=2或-2.2.(2022·山东淄博)化简a2a-1-1-2a1-a的结果为(B)A.a+1a-1B.a-1C.aD.1【解析】原式=a2a-1+1-2aa-1=(a-1)2a-1=a-1.3.(2022·四川南充)已知1x-1y=3,则代数式2x+3xy-2yx-xy-y的值是(D)A.-72B.-112C.92D.34【解析】∵1x-1y=3,∴y-xxy=3,∴x-y=-3xy,则原式=2(x-y)+3xy(x-y)-xy=-6xy+3xy-3xy-xy=-3xy-4xy=34.4.(2022·江苏苏州)计算1+1x÷x2+2x+1x的结果是(B)A.x+1B.1x+1C.xx+1D.x+1x【解析】原式=xx+1x÷(x+1)2x=x+1x·x(x+1)2=1x+1.5.若分式2aba-b中的a,b的值同时扩大到原来的3倍,则分式的值(B)A.不变B.是原来的3倍5\nC.是原来的6倍D.是原来的9倍【解析】若分式2aba-b中的a,b的值同时扩大到原来的3倍,即2×3a×3b3a-3b=6aba-b=3×2aba-b,所以分式的值是原来的3倍.6.下列分式是最简分式的是(B)A.1-xx-1B.2xx2+1C.x-1x2-1D.42x【解析】1-xx-1=-x-1x-1=-1,故A错误;x-1x2-1=x-1(x+1)(x-1)=1x+1,故C错误;42x=2x,故D错误.7.(2022·北京)如果a-b=23,那么代数式a2+b22a-b·aa-b的值为(A)A.3B.23C.33D.43【解析】原式=a2+b22a-2ab2a·aa-b=(a-b)22a·aa-b=a-b2,当a-b=23时,原式=232=3.8.已知14m2+14n2=n-m-2,则1m-1n的值等于(C)A.1B.0C.-1D.-14【解析】由14m2+14n2=n-m-2,得(m+2)2+(n-2)2=0,则m=-2,n=2,∴1m-1n=-12-12=-1.9.(2022·沈阳)化简:2aa2-4-1a-2= 1a+2　. 【解析】原式=2a(a+2)(a-2)-a+2(a+2)(a-2)=a-2(a+2)(a-2)=1a+2.10.(2022·黑龙江大庆)已知3x-4(x-1)(x-2)=Ax-1+Bx-2,则实数A=　1　. 【解析】Ax-1+Bx-2=A(x-2)(x-1)(x-2)+B(x-1)(x-1)(x-2)=(A+B)x-(2A+B)(x-1)(x-2),∵3x-4(x-1)(x-2)=Ax-1+Bx-2,∴A+B=3,2A+B=4,解得A=1,B=2.11.(8分)先化简,再求值:1a-2+1÷a2-13a-6,其中a=tan60°-|-1|.解:∵a=tan60°-|-1|,∴a=3-1,∴原式=a-1a-2·3(a-2)(a+1)(a-1)=3a+1=3.12.(10分)某学生化简分式1x+1+2x2-1出现了错误,解答过程如下:原式=1(x+1)(x-1)+2(x+1)(x-1)(第一步)=1+2(x+1)(x-1)(第二步)5\n=3x2-1.(第三步)(1)该学生解答过程是从第　　　　步开始出错的,其错误原因是　; (2)请写出此题正确的解答过程.解:(1)一;分式的基本性质用错.(2)原式=x-1(x+1)(x-1)+2(x+1)(x-1)=x+1(x+1)(x-1)=1x-1.13.(10分)请你先化简a2a+2-a+2÷4aa2-4,再从-2,2,2中选择一个合适的数代入求值.解:a2a+2-a+2÷4aa2-4=a2a+2-(a-2)(a+2)a+2×(a+2)(a-2)4a=4a+2×(a+2)(a-2)4a=a-2a,为使分式有意义,a不能取±2,所以a=2,当a=2时,原式=2-22=1-2.14.(10分)先化简,再求值:a-1a2-4a+4-a+2a2-2a÷4a-1,其中a是满足不等组7-a>2,2a>3的整数解.解:a-1a2-4a+4-a+2a2-2a÷4a-1=a(a-1)-(a+2)(a-2)a(a-2)2·a4-a=4-aa(a-2)2·a4-a=1(a-2)2,∵解不等式组得32<a<5,∴a=2,3,4,∵原式中a≠0,2,4,∴a=3,当a=3时,原式=1(3-2)2=1.[名师预测]1.若分式x-4x+4无意义,则x的值是(B)5\nA.x=4B.x=-4C.x=0D.x≠-4【解析】∵分式x-4x+4无意义,∴x+4=0,解得x=-4.2.下列分式:a33x2,x-yx2+y2,m2+n2m2-n2,m+1m2-1,a2-2ab+b2a2-2ab-b2,其中最简分式有(C)A.2个B.3个C.4个D.5个【解析】a33x2,x-yx2+y2,m2+n2m2-n2,a2-2ab+b2a2-2ab-b2这4个分式是最简分式;而m+1m2-1=m+1(m+1)(m-1)=1m-1.故最简分式有4个.3.已知x2-3x-4=0,则代数式xx2-x-4的值是(D)A.3B.2C.13D.12【解析】将已知等式整理为x-4x=3,则xx2-x-4=1x-4x-1=13-1=12.4.如果a+b=2,那么代数式a-b2a÷a-ba的值是　2　. 【解析】当a+b=2时,原式=a2-b2a·aa-b=(a-b)(a+b)a·aa-b=a+b=2.5.化简:1+1x-1÷x2+xx2-2x+1= x-1x+1　. 【解析】1+1x-1÷x2+xx2-2x+1=x-1+1x-1·(x-1)2x(x+1)=xx-1·(x-1)2x(x+1)=x-1x+1.6.先化简,再求值:1-1x+1÷xx2-1,其中x=2+1.解:当x=2+1时,原式=xx+1·(x+1)(x-1)x=x-1=2.7.先化简xx-2-4x2-2x÷x+2x2-x,然后从-2,-1,0,1,2中选取一个你喜欢的值代入求值.解:原式=xx-2-4x(x-2)÷x+2x(x-1)=x2-4x(x-2)·x(x-1)x+2=(x+2)(x-2)x(x-2)·x(x-1)x+2=x-1.∵x=-2,0,1,2能使分母为0,无意义,∴x只能取-1,当x=-1时,原式=-1-1=-2.8.先化简,再求值:x2-1x2-4÷-3x-2-x-2,其中x是满足不等式-12(x-1)≥12的非负整数解.解:∵-12(x-1)≥12,∴x≤0,即非负整数解为0,∴x=0,原式=(x+1)(x-1)(x+2)(x-2)÷-3x-2-(x+2)(x-2)x-2=(x+1)(x-1)(x+2)(x-2)×x-2(1-x)(1+x)=-1x+2=-12.9.观察下列等式:5\n第一个等式:a1=21+3×2+2×22=12+1-122+1;第二个等式:a2=221+3×22+2×(22)2=122+1-123+1;第三个等式:a3=231+3×23+2×(23)2=123+1-124+1;第四个等式:a4=241+3×24+2×(24)2=124+1-125+1;…按上述规律,回答下列问题:(1)用含n的代数式表示第n个等式an;(2)求a1+a2+a3+a4+a5+a6的最简结果;(3)计算:a1+a2+…+an.解:(1)an=2n1+3×2n+2×(2n)2=12n+1-12n+1+1.(2)a1+a2+a3+a4+a5+a6=12+1-122+1+122+1-123+1+123+1-124+1+124+1-125+1+125+1-126+1+126+1-127+1=12+1-127+1=1443.(3)a1+a2+…+an=12+1-122+1+122+1-123+1+…+12n+1-12n+1+1=12+1-12n+1+1=2n+1-23(2n+1+1).5