2011年全国各地100份中考数学试卷分类汇编第7章分式与分式方程一、选择题1.(2011浙江金华,7,3分)计算1a-1–aa-1的结果为()A.1+aa-1B.-aa-1C.-1D.1-a【答案】C2.(2011山东威海,8,3分)计算:211(1)1mmm的结果是()A.221mmB.221mmC.221mmD.21m【答案】B3.(2011四川南充市,8,3分)当8、分式21xx的值为0时,x的值是()(A)0(B)1(C)-1(D)-2【答案】B4.(2011浙江丽水,7,3分)计算1a-1–aa-1的结果为()A.1+aa-1B.-aa-1C.-1D.1-a【答案】C5.(2011江苏苏州,7,3分)已知2111ba,则baab的值是A.21B.-21C.2D.-2【答案】D6.(2011重庆江津,2,4分)下列式子是分式的是()A.2xB.1xxC.yx2D.3x【答案】B.7.(2011江苏南通,10,3分)设m>n>0,m2+n2=4mn,则22mnmn的值等于A.23B.3C.6D.3【答案】A8.(2011山东临沂,5,3分)化简(x-x1-x2)÷(1-x1)的结果是()A.x1B.x-1C.x1-xD.1-xx【答案】B9.(2011广东湛江11,3分)化简22ababab的结果是A新疆源头学子小屋特级教师王新敞@126.comwxckt@126.com王新敞特级教师源头学子小屋新疆abB新疆源头学子小屋特级教师王新敞@126.comwxckt@126.com王新敞特级教师源头学子小屋新疆abC新疆源头学子小屋特级教师王新敞@126.comwxckt@126.com王新敞特级教师源头学子小屋新疆22abD新疆源头学子小屋特级教师王新敞@126.comwxckt@126.com王新敞特级教师源头学子小屋新疆1【答案】A二、填空题1.(2011浙江省舟山,11,4分)当x时,分式x31有意义.【答案】3x2.(2011福建福州,14,4分)化简1(1)(1)1mm的结果是.【答案】m3.(2011山东泰安,22,3分)化简:(2xx+2-xx-2)÷xx2-4的结果为。【答案】x-64.(2011浙江杭州,15,4)已知分式235xxxa,当x=2时,分式无意义,则a=,当a6时,使分式无意义的x的值共有个.【答案】6,25.(2011浙江湖州,11,4)当x=2时,分式11x的值是【答案】16.(2011浙江省嘉兴,11,5分)当x时,分式x31有意义.【答案】3x7.(2011福建泉州,14,4分)当x=时,分式22xx的值为零.【答案】2;8.(2011山东聊城,15,3分)化简:2222222ababaabbab=__________________.【答案】9.(2011四川内江,15,5分)如果分式23273xx的值为0,则x的值应为.【答案】-310.(2011四川乐山11,3分)当x=时,112x【答案】311.(2011四川乐山15,3分)若m为正实数,且13mm,221mm则=【答案】13312.(2011湖南永州,5,3分)化简aaa111=________.【答案】1.13.(2011江苏盐城,13,3分)化简:x2-9x-3=▲.【答案】x+3三、解答题1.(2011安徽,15,8分)先化简,再求值:12112xx,其中x=-2.【答案】解:原式=112111)1)(1(1)1)(1(21xxxxxxx.2.(2011江苏扬州,19(2),4分)(2)xxx1)11(2【答案】(2)解:原式=xxxxx)1)(1(1=)1)(1(1xxxxx=11x3.(2011四川南充市,15,6分)先化简,再求值:21xx(xx1-2),其中x=2.【答案】解:方法一:21(2)1xxxx=221211xxxxxx=12(1)(1)(1)(1)xxxxxxxx=121(1)(1)xxxx=12(1)(1)(1)(1)xxxxxx=12(1)(1)xxxx=121(1)(1)(1)(1)xxxxxxx=(1)(1)(1)xxx=11x当x=2时,11x=121=-1方法二:21(2)1xxxx=212()1xxxxxx=2121xxxxx=1(1)(1)xxxxx=(1)(1)(1)xxxxx=11x当x=2时,11x=121=-1.4.(2011浙江衢州,17(2),4分)化简:3abababab.【答案】原式3222()2ababababababab5.(2011四川重庆,21,10分)先化简,再求值:(x-1x-x-2x+1)÷2x2-xx2+2x+1,其中x满足x2-x-1=0.【答案】原式=(x-1x-x-2x+1)÷2x2-xx2+2x+1=(x-1)(x+1)-x(x-2)x(x+1)÷2x2-xx2+2x+1=2x-1x(x+1)×(x+1)22x-1=x+1x2当x2-x-1=0时,x2=x+1,原式=1.6.(2011福建泉州,19,9分)先化简,再求值2221xxxxx,其中2x.【答案】解:原式2(1)(1)(1)xxxxxx··························································4分11x·········································································································6分当2x时,原式1.····················································································9分7.(2011湖南常德,19,6分)先化简,再求值.221211,2.111xxxxxxx其中【答案】解:221211111xxxxxx21111111111122==2.21xxxxxxxxxxxxx当时,原式8.(2011湖南邵阳,18,8分)已知111x,求211xx的值。【答案】解:∵111x,∴x-1=1.故原式=2+1=39.(2011广东株洲,18,4分)当2x时,求22111xxxx的值.【答案】解:原式=2221(1)111xxxxxx当2x时,原式1211x10.(2011江苏泰州,19(2),4分)abababba)2﹢﹣(【答案】(2)原式=abababbababa]))(([2=abababba222=ababaa2=a11.((2011山东济宁,16,5分)计算:22()ababbaaa【答案】原式=222abaabbaa………………2分=2()abaaab………………4分=1ab………………5分12.(2011四川广安,22,8分)先化简22()5525xxxxxx,然后从不等组23212xx≤的解集中,选取一个你认为符合题意....的x的值代入求值.【答案】解:原式=2(5)(5)52xxxxx=5x解不等组得:-5≤x<6选取的数字不为5,-5,0即可(答案不唯一)13.(2011重庆江津,21(3),6分)先化简,再求值:)121(212xxx,其中31x·【答案】(3)原式=2212)1)(1(xxxxx=)1(22)1)(1(xxxxx=1-x·把31x代入得原式=1-31=32·14.(2011江苏南京,18,6分)计算221()abababba【答案】221)abababba解:(()()()()aabbababababba()()bbaababb1ab15.(2011贵州贵阳,16,8分)在三个整式x2-1,x2+2x+1,x2+x中,请你从中任意选择两个,将其中一个作为分子,另一个作为分母组成一个分式,并将这个分式进行化简,再求当x=2时分式的值.【答案】解:选择x2-1为分子,x2+2x+1为分母,组成分式x2-1x2+2x+1.x2-1x2+2x+1=(x+1)(x-1)(x+1)2=x-1x+1.将x=2代入x-1x+1,得13.16.(2011广东肇庆,19,7分)先化简,再求值:)211(342aaa,其中3a.【答案】解:)211(342aaa=)2122(3)2)(2(aaaaaa=233)2)(2(aaaaa=2a当3a时,原式=2a=12317.(20011江苏镇江,18(1),4分)(2)化简:22142xxx答案:(2)原式=222222221222xxxxxxxxxxx18.(2011重庆市潼南,21,10分)先化简,再求值:2121(1)1aaaa,其中a=2-1.【答案】解:原式=211(1)1aaaa---------------------4分=1a----------------------8分当a=2时,原式=2112---------------------10分19.(2011山东枣庄,19,8分)先化简,再求值:1+1x-2÷x2-2x+1x2-4,其中x=-5.解:412)211(22xxxx=)2)(2()1(2122xxxxx……………………2分=2)1()2)(2(21xxxxx=12xx,………………………………………………………………………5分当5x时,原式=12xx=211525.………………………………………8分20.(2011湖北宜昌,16,7分)先将代数式11)(2xxx化简,再从-1,1两数中选取一个适当的数作为x的值代入求值.【答案】解:原式=11)1(xxx=x(3分,省略不扣分)=x(6分)当x=1时,原式=1.(7分)(直接代入求值得到1,评4分),