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一、选择题1.（房山区二模）1.若代数式22xx有意义，则实数x的取值范围是A.0xB．2xC．0xD．2x答案：D2.（海淀区二模）1．若代数式31x有意义，则实数x的取值范围是A.1xB.1xC.1xD.0x答案：C

3.（朝阳区二模）1.若代数式3xx的值为零，则实数x的值为（A）x=0（B）x≠0（C）x=3（D）x≠3答案：A4.（丰台区二模）6．已知111mn，则代数式222mmnnmmnn的值为（A）3（B）1（A）-1（D）-3答案:

D5.．（市朝阳区一模）若代数式32x有意义，则实数x的取值范围是（A）x=0（B）x=3（C）x≠0（D）x≠3答案D6．（延庆区初三统一练习）计算：97...aaabbbb个个A．97abB．97abC．79abD．97ab答案：C7．（海淀区第二学期练习）如果1ab

，那么代数式2222(1)baaab的值是A.2B.2C.1D.1答案A8.（怀柔区一模）若代数式3-xx2有意义，则实数x的取值范围是（）A.x=0B.x≠3C.x≠0D.x=3答案B9．（门头沟区初三综合练习）如果代数式3xx有意义

，则实数x的取值范围是A．3x≥B．0xC．30xx≥-且D．3x≥答案C10.（房山区一模）如果30ab，那么代数式2222()abbabaaa的值是A.12B.1-2C.14D.1答案A11.（昌平区初二年级期末）如果分式33x在实数范围

内有意义，那么x的取值范围是A.x＜-3B．x＞-3C．x≠-3D．x=-3答案：C12．（市朝阳区初二年级第一学期期末）若分式21xx的值为0，则实数x的值为A．2B．1C．0D．1答案：A13．（市东城区初二期末）若分式23xx的值为0，则x的值等于A．0B．2C．3D．-3

解：B14．（市丰台区初二期末）一件工作，甲单独完成需要a天，乙单独完成需要b天，如果甲、乙二人合作，那么每天的工作效率是A．a+bB．1a+1bC．1a+bD．aba+b答案：B15．（市海淀区八年级期末）若分式

1aa的值等于0，则a的值为A．1B．1C．2D．2答案：A16．（市怀柔区初二期末）如果分式+1xx的值为零，那么x的值为A.0B.1C.-1D.1答案：A17．（市门头沟区八年级期末）如果将分式2xxy中的字母x与y的值分别扩大为原来的10倍，那么这个分式

的值A．扩大为原来的10倍B．扩大为原来的20倍C．缩小为原来的110D．不改变答案：D18．（市平谷区初二期末）下列各式中，正确的是A．326xxxB．nmnxmxC．1112xxxD．1yxyx答案：C19．（市石景山区初二期末）如果12ab，那么代数

式2baaaab的值是A．2B．2C．12D．12答案：D20．（市石景山区初二期末）当分式623x的值为正整数时，整数．．x的取值可能有A．4个B．3个C．2个D．1个答案：C21．（市顺义区八年级期末）下

列各式从左到右的变形正确的是A．yxyx=－1B．yx=11yxC．yxx=y11D．2)3(yx=223yx答案：A22．（市西城区八年级期末）化简分式277()abab的结果是（）．A．7abB．7abC．7abD．7ab答案：B23．（市西城区八年级

期末）已知12xy，则3xyy的值为（）．A．7B．17C．52D．25答案：C24（大兴区八年级第一学期期末）如果分式12+2x有意义，那么x的取值范围是A.0xB.-1xC.1xD.1x25.（大兴区八年级第一学期

期末）5.如果将分式yxy2（x,y均为正数）中字母的x，y的值分别扩大为原来的3倍，那么分式yxy2的值A．扩大为原来的3倍B．不改变C．缩小为原来的13D．扩大为原来的9倍26.（延庆区八年级第一学区期末）若代数式4xx有意义，则实数x的取值范围是A．0xB．4x

C．0xD．4x答案：D27.（延庆区八年级第一学区期末）如果把yxy322中的x和y都扩大5倍，那么分式的值A．扩大5倍B．不变C．缩小5倍D．扩大4倍答案：B28．（市西城区八年级期末）要使分式21x有意义，则x的取值范围是．答案：x≠129.（市门头

沟区八年级期末）如果分式21xx的值为0，那么x=．答案：230.（昌平区初二年级期末）如果分式241xx的值为0，那么x的值为.答案：231．（市海淀区八年级期末）已知分式满足条件“只含有字母x，且当x=1时无意义”，请写出一个这样

的分式：．答案：11x32．（市怀柔区初二期末）若代数式xx-4有意义，则x的取值范围是____________．答案：x433．（市门头沟区八年级期末）学习了“分式的加法”的相关知识后，小明同学画出了下图：请问他画的图中①为，②为．答案：略34.（

市平谷区初二期末）若分式12xx值为0，则x的值是________.解：235．（市石景山区初二期末）分式变形224xAxx中的整式A=，变形的依据是．解：22xx，分式的基本性质36．（市石景山区初二期末）计

算238932xyyx=．解：212yx37．（市顺义区八年级期末）已知1132ab，则代数式60°DCAB254436aabbabab的值为．答案：1238．（市西城区八年级期末）计算：（1）223()ba=________

______；（2）21054abacc=______________．答案：（1）429ba；（2）8bc．（各2分）39.（东城区二模）若分式22xx的值为正，则实数x的取值范围是__________________.答案：x＞040.（燕山地区

一模）当a=3时，代数式212)212(22aaaaaa的值是．答案：241.（海淀区二模）11．如果3mn，那么代数式nmmmnnm的值是．答案：442．（燕山地区一模）如果分式4xx的

值是0，那么x的值是．答案：x=0．43．（延庆区初三统一练习）如果210aa，那么代数式221()1aaaaa的值是．答案：144．（延庆区初三统一练习）若分式23xx有意义，则实数x的取值范围是．答案：x≠345．（西城区

九年级统一测试）若代数式11xx的值为0，则实数x的值为__________．答案：146.（昌平区二模）如果230aa，那么代数式221()1aaaaa的值是．答案：347．（顺义区初三练习）如果2240nn，

那么代数式242nnnn的值为．答案：448.（通州区一模）答案：349．（石景山区初三毕业考试）如果5xy，那么代数式221+yxxyxy()的值是_______．答案：5

50.（怀柔区一模）如果x+y-1=0,那么代数式xyxxyx2的值是__________.答案151..（门头沟区初三综合练习）如果23ab，那么22242abaab的结

果是．答案452．（市大兴区检测）23yx，则222569222yxxyyxyxyxy的值是.答案353．（丰台区一模）如果代数式221mm，那么22442mmmmm的值为．答案154.（东城区一模）化简代数式11+12

2xxxx，正确的结果为________________.答案2x55.（市朝阳区综合练习（一））如果023nm，那么代数式)2(4322nmnmnm的值是．答案4756．（市朝阳区

一模）先化简，再求值：1111122aaaaa，其中4a．解：1111122aaaaa=11)1()1)(1(2aaaaaa„„„„„„„„„………………2分11a．………………………

„„…………………………………4分当4a时，原式=51.…………………………………………………………………5分57.（门头沟区第一学期期末调研试卷）20.先化简，再求值：2211mmmmm，其中m是方程230xx的根．答

案：原式=22211mmmmm=22(1)1mmmm=2mm．„„„„„„3分∵m是方程230xx的根，∴230mm．∴23mm．„„„„„„„„„5分58.（昌平区初二年级期末）计算：22142aaa．解：

原式=21(+2)(-2)2aaaa„„„„„„„„„„„„„„1分=22(+2)(-2)22aaaaaa„„„„„„„„2分=2-(+2)(+2)(-2)aaaa„„„„„„„„„„„„3分=-2(+2)(-2)aaa„„„„„„„„„„„„

„4分=1+2a„„„„„„„„„„„„„„„„„„„„„„„5分59.（昌平区初二年级期末）先化简，再求值：22121211xxxxx，其中3x．解：原式=12121122xxxxx„„„„„„„„„„„„„„„„1分=12)1()1)(112xxx

xx（„„„„„„„„„„„„„„„„„„„2分=121)1xxxx（„„„„„„„„„„„„„„„„„„„„„„„3分=121)(1)xxxxxx（„„„„„„„„„„„„„„„„„4分=x1.„„„„„„„„„„„„„„„„„„„„„„„„„„5分当3x时

，原式=33„„„„„„„„„„„„„„„„„„„„„„6分60．（市朝阳区初二年级第一学期期末）计算：4222xxxxxx．解：4222xxxxxx(2)(2)2(2)(2)4xxxxxxxx„„„„

„„„„„„„„„„2分42(2)(2)4xxxxx„„„„„„„„„„„„„„„„3分12x．„„„„„„„„„„„„„„„„„„„„„„„„4分61．（市朝阳区初二年级第一学期期末）分式中，在分

子、分母都是整式的情况下，如果分子的次数低于分母的次数，称这样的分式为真分式．例如，分式42x，2334xxx是真分式．如果分子的次数不低于分母的次数，称这样的分式为假分式．例如，分式11xx，21xx是假分式．一个假分式可以化为一个整式与一个真分

式的和．例如，1(1)221111xxxxx．（1）将假分式211xx化为一个整式与一个真分式的和；（2）若分式21xx的值为整数，求x的整数值．解：（1）213211xxx．„„

„„„„„„„„„„„„„„„„„„„„2分（2）21111xxxx．„„„„„„„„„„„„„„„„„„„„4分∵分式21xx的值为整数，且x为整数，∴11x或11x．解得0x或2x．„„„„„„„„„„„„„„„„6分62.（市东

城区初二期末）先化简，再求值：259123xxx，其中32x．【解析】333223333233142xxxxxxxxxxx解：原式分分分当32x时，原式11333223．…5

分63．（市丰台区初二期末）计算：1-1m-2æèçöø÷¸m-32m-4．答案：64．（市丰台区初二期末）先化简，再求值：xxxx2393131，其中x=3-3.答案：65．（市海淀区八年级

期末）先化简，再求值：2442()mmmmm，其中3m．解：原式=22442mmmmm-------------------------------------------1分=22442mmmmm=2222mmmm----

--------------------------------------------2分=22mm．-------------------------------------------------3分当3m时，原式=15．---------------------

------------------------5分66．（市怀柔区初二期末）计算：22yxxyyxy．解：原式=()()()yxyxyxyxy(x+y)(x-y)„„„„„„„2分=2yxyxy(x+y)(x-y)„„„„„„„4分=222yxy„„„„„„„5

分67．（市怀柔区初二期末）先化简：22211aaaaaa，然后从0，1，2中选一个你认为合适的a值，代入求值.解：原式＝a2－2a＋1a÷1－a2a2＋a„„„„„„„1分＝(a－1)2a·a(a＋1)(1－a)

(a＋1)„„„„„„„3分＝1－a„„„„„„„4分当a=2时，原式=1－a=1-2=-1„„„„„„„5分68．（市门头沟区八年级期末）已知30ab，求222ababaabb的值．解：222ababaabb2ababab„„„

„„„„„„„„„„„„„„„„„„„„„„„2分.abab„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分当30ab，即3ab时，原式31.32abbbabbb„„„„„„„„„„„„„„5分69．（市门头沟区八年级

期末）阅读材料，并回答问题：小明在学习分式运算过程中，计算1122xx的解答过程如下：解：1122xx①222222xxxxxx②22xx③22xx④4.

⑤问题：（1）上述计算过程中，从步开始出现了错误（填序号）；（2）发生错误的原因是：；（3）在下面的空白处，写出正确的解答过程：解：（1）从第③步开始出现错误；„„„„„„„„„„„„„„„„„„„„„1分（2）略；„„„„„„„„„„„„„„„„„„„„„„„

„„„„„„„2分（3）1122xx222222xxxxxx2222xxxx„„„„„„„„„„„„„„„„„„„„„„„„„3分422xx„

„„„„„„„„„„„„„„„„„„„„„„„„„4分24.4x„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分70．（市平谷区初二期末）计算：xxx31922解：原式313

32xxxx„„„„„„„„„„„„„„„„1333332xxxxxx„„„„„„„„„„„„„23332xxxx3332xxxx„„„„„„„„„„„„„„„„3333x

xx„„„„„„„„„„„„„„„„„431x„„„„„„„„„„„„„„„„„„„571．（市平谷区初二期末）已知：0232aa，求代数252232aaaaa的值．解：原式

25222232aaaaaaa„„„„„„„„„„„„„„„12542322aaaaa„„„„„„„„„„„„„„„„„233223aaaaaa„„„„„„„„„„„„„„„331aa„„„„„„„

„„„„„„„„„„„„„„„„40232aa232aa原式21312aa„„„„„„„„„„„„„„„„„„„572.（市平谷区初二期末）若311ba，求bababa22的值.解：311baabab3„„„„„„„„„„„„„„.

.2bababa22abbaba)(2„„„„„„„„„„„„„„„„„3ababab63„„„„„„„„„„„„„„„„„453„„„„„„„„„„„„„„„„„573．（市石景山区初二期末）当21x时，求代数式21112441xx

xxxx的值．解：原式22112(1)1xxxxx⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯1分211(1)xxxx⋯⋯⋯⋯⋯⋯⋯⋯⋯2分2(1)(1)xxx⋯⋯⋯⋯⋯⋯⋯⋯⋯3分11x⋯⋯⋯⋯

⋯⋯⋯⋯⋯4分当21x时，原式122211⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯5分74.（市顺义区八年级期末）（4分）已知：1xy，3(2)64xy，求代数式22xyxy的值.解：∵1xy，3(2)64xy

，∴124xyxy„„„„„„„„„„„„„„„„„„2分（各1分）解得21xy„„„„„„„„„„„„„„„„„4分（各1分）∴2222213215xyxy„„„„„„„„„„„„„„„5分75.（市顺义区八年级期末）（5分）先化简，再求值：2532

236xxxxx，其中x满足2310xx.解：原式=(2)(2)5323(2)xxxxxx„„„„„„„„„1分=293(2)23xxxxx„„„„„„„„„„„„„„„„„2分=(3)(3

)3(2)23xxxxxx„„„„„„„„„„„3分=239xx„„„„„„„„„„„„„„„„„4分∵2310xx∴231xx∴原式=22393(3)313xxxx„„„„„„„

„5分76.（市西城区八年级期末）老师所留的作业中有这样一个分式的计算题：22511xxx，甲、乙两位同学完成的过程分别如下：老师发现这两位同学的解答都有错误．请你从甲、乙两位同学中，选择一位同学的解答过程，帮助他分析错

因，并加以改正．（1）我选择________同学的解答过程进行分析．（填“甲”或“乙”）该同学的解答从第________步开始出现错误，错误的原因是_______________________________________________________________________

_____________________；（2）请重新写出完成此题的正确解答过程．22511xxx甲同学：22511xxx=25(1)(1)(1)(1)xxxxx第一步=25(1)(1)xxx第二步=7(1)(1)xx

x第三步乙同学：22511xxx=2(1)5(1)(1)(1)(1)xxxxxx第一步=225xx第二步=33x第三步解：（1）选甲：一，理由合理即可，如：第一个分式的变形不符合分式的基本性质，分子漏乘1x；„„„„„„„„„„„„„„„„„

„„„„„„„„2分选乙：二，理由合理即可，如：与等式性质混淆，丢掉了分母；„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分（2）22511xxx=2(1)5(1)(1)(1)(1)xxxxxx„„„„„„„„„„„

„„„„„„„„3分=225(1)(1)xxxx=33(1)(1)xxx„„„„„„„„„„„„„„„„„„„„„„„„„4分=31x．„„„„„„„„„„„„„„„„„„„„„„„5分77.（延庆区八年级第一学区期末）李老

师在黑板上写了一道题目，计算：23311xxx.小宇做得最快，立刻拿给李老师看，李老师看完摇了摇头，让小宇回去认真检查.请你仔细阅读小宇的计算过程，帮助小宇改正错误．23311xxx=33111xxxx（A）=

3131111xxxxxx（B）=33(1)xx（C）=26x（D）（1）上述计算过程中，哪一步开始．．出现错误？；(用字母表示)（2）从（B）到（C）是否正确？；若不正确，错误的原因是；（3）请你写出

此题完整正确的解答过程．.(1)A„„„„„„„„„„1分(2)否,根据分式加减法法则：同分母分式相加减，分母不变，分子相加减，小宇把分母去掉．．．．．了．„„„„„„„„„3分(3)解：„„„„„„„„„„4分„„„„„„„„„„5分78..（延庆区八年级第一学区期末）先化简，再求

值：121112aaaa，其中13a．解：„„„„„„„„„„2分„„„„„„„„„„„„„3分„„„„„„„„„„„„„„„„„4分当13a时，原式„„„„„„„„„„„„5分