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2022年中考数学专题04分式及其运算试题 .pdf

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1、专题 04 分式及其运算?解读考点知识点名师点晴分式的概念整式A除以整式B，可以表示成AB的形式，如果除式B中含有字母，那么称AB为分式若B0，则AB有意义；若B=0，则AB无意义；若A=0且B 0，则AB0.分式的基本性质及应用1.分式的基本性质)0()0(CCBCABACCBCABA要熟练掌握，特别是乘或除以的数不能为0 2.分式的变号法则分式的分子、分母与分式本身的符号，改变其中任何两个，分式的值不变.3.分式的约分、通分通分与约分的依据都是分式的基本性质4.最简分式分子与分母没有公因式分式的运算1.分式的加减法异分母的分式相加减，要先通分，然后再加减2.分式的乘除法、乘方熟练应用法则

2、进行计算3.分式的混合运算应先算乘方，再将除法化为乘法，进行约分化简，最后进行加减运算若有括号，先算括号里面的灵活运用运算律，运算结果必须是最简分式或整式?2 年中考【20XX年题组】1（2015 常州）要使分式23x有意义，则x的取值范围是（）A2x B2x C2x D2x【答案】D【解析】试题分析：要使分式23x有意义，须有20 x，即2x，故选 D考点：分式有意义的条件2（2015 济南）化简2933mmm的结果是（）A3m B3m C33mm D33mm【答案】A 考点：分式的加减法3（2015 百色）化简222624xxxxx的结果为（）A214x B212xx C12x D62x

3、x【答案】C【解析】试题分析：原式=262(2)(2)xxxx=2(2)(6)(2)(2)xxxx=2(2)(2)xxx=12x故选 C考点：分式的加减法4（2015 甘南州）在盒子里放有三张分别写有整式a+1，a+2，2 的卡片，从中随机抽取两张卡片，把两张卡片上的整式分别作为分子和分母，则能组成分式的概率是（）A13 B23 C16 D34【答案】B【解析】试题分析：分母含有字母的式子是分式，整式a+1，a+2，2 中，抽到a+1，a+2 做分母时组成的都是分式，共有 32=6 种情况，其中a+1，a+2 为分母的情况有4 种，所以能组成分式的概率=46=23故选 B考点：1概率公式；2分

4、式的定义；3综合题5（2015 龙岩）已知点P（a，b）是反比例函数1yx图象上异于点（1，1）的一个动点，则1111ab=（）文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 Z

5、O6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编

6、码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P

7、2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6

8、 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文

9、档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A1

10、0P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9文档编码:CW1V3A9A10P2 HG5E5H6N2K6 ZO6J7W7H4N9A2 B1 C32 D12【答案】B 考点：1反比例函数图象上点的坐标特征；2分式的化简求值；3条件求值6（2015 山西省）化简222

11、22aabbbabab的结果是（）Aaab Bbab Caab Dbab【答案】A【解析】试题分析：原式=2()()()abbab abab=abbabab=abbab=aab，故选 A考点：分式的加减法7（2015 泰安）化简：341()(1)32aaaa的结果等于（）A2a B2a C23aa D32aa【答案】B【解析】试题分析：原式=(3)342132a aaaaa=24332aaaa=(2)(2)332aaaaa=2a故选 B 考点：分式的混合运算8（2015 莱芜）甲乙两人同时从A地出发到B地，如果甲的速度v保持不变，而乙先用12v的速度到达中点，再用2v的速度到达B地，则下列结论

12、中正确的是（）A甲乙同时到达B地 B甲先到达B地C乙先到达B地 D谁先到达B地与速度v有关【答案】B 文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9

13、U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9

14、 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9

15、U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9

16、 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9

17、U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9

18、 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3考点：1列代数式（分式）；2行程问题9（2015 内江）已知实数a，b满足：211aa，211bb，则2015a b|=【答案】1【解析】试题分析：2110aa，2110bb，0a，0b，()10ab ab，211aa，211bb，两式相减可得2

19、211abab，()()baab abab，()1()0ab abab，0ab，即ab，2015a b=02015=1故答案为：1考点：1因式分解的应用；2零指数幂；3条件求值；4综合题；5压轴题10（2015 黄冈）计算)1(22baabab的结果是 _【答案】1ab【解析】试题分析：原式=()()babaab abab=()()babababb=1ab故答案为：1ab考点：分式的混合运算11（2015 安徽省）已知实数a、b、c满足ababc，有下列结论：若c0，则111ab；若a3，则bc9；若abc，则abc0；若a、b、c中只有两个数相等，则abc8其中正确的是（把所有正确结论的序号

20、都选上）【答案】文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6

21、C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1

22、F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6

23、C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1

24、F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6

25、C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1

26、F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3考点：1分式的混合运算；2解一元一次方程12（2015 梅州）若1212)12)(12(1nbnann，对任意自然数n都成立，则a，b；计算：21191751531311m【答案】12；12；1021【解析】试题分析：1(21)(21)nn=2121abnn=(21)(21)(21)(21)anbnnn=(22)(21)(21)ab n

27、abnn，可得(22)1ab nab，即：01abab，解得：a=12，b=12；m=111111(1.)23351921=11(1)221=1021，故答案为：12；12；1021考点：1分式的加减法；2综合题13（2015 河北省）若02ba，则ababa222的值为【答案】32【解析】试题分析：2ab，原式=2222442bbbb=32，故答案为：32考点：分式的化简求值文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9

28、ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U

29、7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9

30、ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U

31、7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9

32、ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U

33、7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9

34、ZZ1F1Y5T5I314（2015 绥化）若代数式25626xxx的值等于0，则x=_【答案】2【解析】试题分析：由分式的值为零的条件得2560 xx，2x 6 0，由2560 xx，得x=2 或x=3，由 2x60，得x3，x=2，故答案为：2考点：分式的值为零的条件15（2015 崇左）化简：2221(1)2aaaa【答案】12a考点：分式的混合运算16（2015 桂林）先化简，再求值：2269392xxxx，其中23x【答案】23x，2【解析】试题分析：分解因式后，利用除法法则变形，约分得到最简结果，把x的值代入计算即可求出值试题解析：原式=2(3)2(3)(3)3xxxx=23x，当

35、23x时，原式=2233=22=2考点：分式的化简求值17（2015 南京）计算：22221()aabaabab【答案】21a【解析】试题分析：首先将括号里面通分运算，进而利用分式的性质化简求出即可文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文

36、档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R

37、2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文

38、档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R

39、2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文

40、档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R

41、2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3试题解析：原式=21()()()abababa aba=2()()()()aababa ab aba ababa=2()()()aababa ababa=21a考

42、点：分式的混合运算18（2015 苏州）先化简，再求值：2121122xxxx，其中31x【答案】11x，33考点：分式的化简求值19（2015 盐城）先化简，再求值：)()(131112aaa，其中a=4【答案】31aa，4【解析】试题分析：根据分式混合运算的法则把原式进行化简，再求出x的值代入进行计算即可试题解析：原式=2113(1)(1)(1)aaaaa=23(1)(1)(1)aaaaa=31aa；当a=4 时，原式=3441=4考点：分式的化简求值20（2015 成都）化简：211()242aaaaa文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编

43、码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G

44、6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编

45、码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G

46、6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编

47、码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G

48、6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编

49、码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3【答案】12aa【解析】试题分析：括号内先通分，同时把除法转化为乘法，再用分式乘法法则计算机即可试题解析：原式=22221212214412212aaaaaaaaaaaaa考点：分式的加减法21（2015 资阳）先化简，再求值：2112()111xxxx，其中x满足260 x【答案】22x，25考点：1分式的混合运算；2分式的化简求值22（2015 达州）化简2221432aaaaaa，并求值，其中a与 2、3 构成ABC的三边，且a为整数【

50、答案】13a，1【解析】试题分析：原式第一项约分后，两项通分并利用同分母分式的减法法则计算得到结果，把a的值代入计算即可求出值试题解析：原式=21(2)(2)(3)2aaaaa aa=11(2)(3)2aaa=13(2)(3)aaa=2(2)(3)aaa=13a，a与 2、3 构成ABC的三边，且a为整数，1a5，即a=2，3，4，当a=2 或a=3 时，原式没有意义，则a=4时，原式=1考点：1分式的化简求值；2三角形三边关系文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码:CQ9U7R6C7U9 HV2R2G6K5V9 ZZ1F1Y5T5I3文档编码

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