【文档说明】(通用版)中考数学一轮复习练习卷1.3《分式》课后练习(含答案).doc,共(8)页,42.000 KB,由MTyang资料小铺上传
转载请保留链接:https://www.ichengzhen.cn/view-35338.html
以下为本文档部分文字说明:
第3节分式(建议答题时间:45分钟)命题点一分式的概念及性质1.下列分式中,是最简分式的是()A.3x24xyB.x2+y2x+yC.x-2x2-4D.1+xx2+2x+12.若代数式xx-4有意义,则实数x的取值范围是()A.x=0B.x=4C.x≠0D.x≠43.若
分式2x-4x+1的值为0,则x的值为________.4.使式子11-2x有意义的x的取值范围为________.命题点二分式化简及求值5.化简:xx-y-yx+y,结果正确的是()A.1B.x2+y2x2-y2C.x-yx+yD.x2+y26.化简4xx2-4-xx-2的结果是(
)A.-x2+2xB.-x2+6xC.-xx+2D.xx-27.若3-2xx-1=()+1x-1,则()中的数是()A.-1B.-2C.-3D.任意实数8.化简(1-2x-1x2)÷(1-1x2)的结果为()A.x-1x+1B.x+1x-
1C.x+1xD.x-1x9.化简:x+3x2-2x+1÷x2+3x(x-1)2=________.10.化简:(1-1a-1)÷a2-4a+4a2-a.11.化简:(a+1a+2)÷(a-2+3a+2).12.计算:x2+4x+4x2-2x÷(x2+x-2x-2-x-2).13.计算:
(3y2x-y-x-y)÷x2-2xyx2-xy.14.计算:x3-x-x2+8x+16x2+3x÷(-2x+3+4x-1).15.计算:x-2x2-2x+1÷(2x-1x-1-x-1)-1x.16.计算:x2-8x+16x2+
2x÷(x-2-12x+2)-1x+4.17.计算:12m÷(m-1+2m+1m+1)-1m.18.先化简,再求值:(mm-2-2mm2-4)÷mm+2,请在2,-2,0,3当中选一个合适的数代入求值.19.先化简,再求值:x+3x
-2÷(x+2-5x-2),其中x=3+3.20.先化简,再求值:(n2n-m-m-n)÷m2,其中m-n=2.21.先化简,再求值:(x-1)÷(2x+1-1),其中x为方程x2+3x+2=0的根.22.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-x≤32x-4<1的整数解中选取.答案1.B2.D3.24.x<125.B6.C7.B8.A9.1x10.解:原式=a-2a-1·a(a-1)(a-2)2=aa-2.11.解:原式=a2+2a+1a
+2÷a2-1a+2=(a+1)2a+2·a+2(a+1)(a-1)=a+1a-1.12.解:原式=(x+2)2x(x-2)÷x2+x-2-(x+2)(x-2)x-2=(x+2)2x(x-2)÷x2+x-2-x2+4x-2=(x+2)2x(x-2)·x-2x+2=x+2x.13.解:原式=3y2-
(x+y)(x-y)x-y·x2-xyx2-2xy=3y2-x2+y2x-y·x(x-y)x(x-2y)=4y2-x2x-y·x-yx-2y=(2y-x)(2y+x)x-y·x-yx-2y=-(2y+x
)=-2y-x.14.解:原式=x3-x-(x+4)2x(x+3)÷-2x+4x+12-x(x+3)x(x+3)=x3-x-(x+4)2x(x+3)÷-x2-x+12x(x+3)=x3-x-(x+4)2x(x+3)·x(x+3)-(x+4)(x-3)=x3-x-x+4-(x-3)=-
43-x.15.解:原式=x-2(x-1)2÷2x-1-(x+1)(x-1)x-1-1x=x-2(x-1)2·x-1-x(x-2)-1x=1-x(x-1)-1x=1-x(x-1)+x-1-x(x-1)=x-x(x-1)=11-x.1
6.解:原式=(x-4)2x(x+2)÷(x-2)(x+2)-12x+2-1x+4=x-4x(x+4)-1x+4=x-4-xx(x+4)=-4x(x+4)=-4x2+4x.17.解:原式=m2÷m2-1+
2m+1m+1-1m=m2·m+1m(m+2)-1m=m+12(m+2)-1m=m(m+1)-2(m+2)2(m+2)m=m2-m-42(m+2)m=m2-m-42m2+4m.18.解:原式=[mm-2-2m(m-2)(m+2)]·m+2m=m+2m-2-2m-2=mm-2,∵m≠±2,0,∴当
m=3时,原式=3.19.解:原式=x+3x-2÷(x2-4x-2-5x-2)=x+3x-2÷x2-9x-2=x+3x-2·x-2x2-9=x+3x-2·x-2(x+3)(x-3)=1x-3,当x=3+3时,原式=13+3-3=13=33.20.解:原式=
[n2n-m-(m+n)]·1m2=n2-n2+m2n-m·1m2=1n-m,∵m-n=2,∴n-m=-2,则原式=1-2=-22.21.解:原式=(x-1)÷2-x-1x+1=(x-1)·x+11-x=-x-1,∵x为方程x2+3x+2=0的根,∴x=-1或x=-2,要使原分式
有意义,则x≠±1,∴x=-2,∴原式=2-1=1.22.解:原式=(x-1)(x+1)+3-3xx+1÷x(x-1)x+1=x2-3x+2x+1·x+1x(x-1)=(x-1)(x-2)x+1·x+1x(x-1)=x-2x,解不等式组2-x≤32x-4<1,得-1≤x<52,∴其整数解
为-1,0,1,2.要使分式有意义,则x不等于-1,0,1,∴x只能取2,当x=2时,原式=0.