2013中考全国100份试卷分类汇编：实数运算

2013中考全国100份试卷分类汇编

实数运算

1、（2013?衡阳）计算

的结果为（ ）

C

2、（2013?常德）计算+的结果为（ ）

=3、（2013年河北）下列运算中，正确的是

Ａ．9＝±3 Ｂ．3－8＝2 Ｃ．(－2)0＝0 D ．2－1

＝12

答案：D

解析：9是9的算术平方根，9＝3，故A 错；3

－8＝－2，B 错，(－2)0＝1，C 也错，选D 。 4、（2013台湾、6）若有一正整数N 为65、104、260三个公倍数，则N 可能为下列何者？（ ）

A ．1300

B ．1560

C ．1690

D ．1800 考点：有理数的混合运算． 专题：计算题．

分析：找出三个数字的最小公倍数，判断即可．

解答：解：根据题意得：65、104、260三个公倍数为1560．

故选B

点评：此题考查了有理数的混合运算，弄清题意是解本题的关键．5、（2013?攀枝花）计算：2﹣1﹣（π﹣3）0﹣=﹣1．

﹣=

．

6、（2013?衡阳）计算=2．

（﹣×=2

7、（2013?十堰）计算：+（﹣1）﹣1+（﹣2）0=2．

．

8、（2013?黔西南州）已知，则a b=1．

9、（2013杭州）把7的平方根和立方根按从小到大的顺序排列为 ． 考点：实数大小比较． 专题：计算题．

分析：先分别得到7的平方根和立方根，然后比较大小． 解答：解：7的平方根为﹣

，

；7的立方根为

， 所以7的平方根和立方根按从小到大的顺序排列为﹣＜

＜

．

故答案为：﹣

＜

＜

．

点评：本题考查了实数大小比较：正数大于0，负数小于0；负数的绝对值越大，这个数越小．

10、（2013?娄底）计算：= 2 ．

×+211、（2013?恩施州）25的平方根是 ±5 ．

12、（2013陕西）计算：=-+-03)13()2( ．

考点：本题经常实数的简单计算、特殊角的三角函数值及零（负）指数幂及绝对值的计算。 解析：原式=718-=+-

13、（2013?遵义）计算：20130﹣2﹣1

= ．

，．故答案为：．14、（2013?白银）计算：2cos45°﹣（﹣）﹣1

﹣﹣（π﹣）0

． 角的余弦等于﹣）×

﹣（﹣﹣．15、（2013?宜昌）计算：（﹣20）×（﹣

1

2

）+．

16、（2013成都市）计算：2

- （2）解析：

（1）2- （2）

=4+23=4

2

17、（2013?黔西南州）（1）计算：

．

×

﹣

18、（2013?荆门）（1）计算：

×

19、（2013?咸宁）（1）计算：+|2﹣|﹣（1

2

）﹣1

+2﹣．

20、（2013?毕节地区）计算：．

21、（2013安顺）计算：2sin60°+2﹣1﹣20130﹣|1﹣|

考点：实数的运算；零指数幂；负整数指数幂；特殊角的三角函数值．

专题：计算题．

分析：本题涉及零指数幂、特殊角的三角函数值、绝对值、负指数幂等四个考点．针对每个考点分别进行计算，然后根据实数的运算法则求得计算结果．

解答：解：原式=2×+﹣1﹣（﹣1）=．

点评：本题考查实数的综合运算能力，是各地中考题中常见的计算题型．解决此类题目的关键是熟记特殊角的三角函数值，熟练掌握零指数幂、特殊角的三角函数值、绝对值、负指数幂等考点的运算．

22、（2013安顺）计算：﹣++= ．

考点：实数的运算．

专题：计算题．

分析：本题涉及二次根式，三次根式化简等考点．针对每个考点分别进行计算，然后根据实数的运算法则求得计算结果．

解答：解：﹣++

=﹣6++3

=﹣．

故答案为﹣．

点评：本题考查实数的综合运算能力，是各地中考题中常见的计算题型．解决此类题目的关键是熟记特殊角的三角函数值，熟练掌握负整数指数幂、零指数幂、二次根式、绝对值等考点的运算．

23、（2013?玉林）计算：+2cos60°﹣（π﹣2﹣1）0．

零指数幂的运算，然后特殊角的三角函数值后合并即可得×﹣

24、（2013?郴州）计算：|﹣|+（2013﹣）0﹣（）﹣1﹣2sin60°．

×

25、（2013?钦州）计算：|﹣5|+（﹣1）2013+2sin30°﹣．

×

26、（2013?湘西州）计算：（）﹣1﹣﹣sin30°．

﹣

﹣

．

27、（13年北京5分14）计算：10)4

1(45cos 22)31(-+?--+-。 解析：

28、（13年山东青岛、8）计算：___________52021=÷+- 答案：

5

2

解析：原式＝

122

+＝52

29、（2013台湾、1）计算12÷（﹣3）﹣2×（﹣3）之值为何？（ ） A ．﹣18 B ．﹣10 C ．2 D ．18 考点：有理数的混合运算． 专题：计算题．

分析：根据运算顺序，先计算乘除运算，再计算加减运算，即可得到结果． 解答：解：原式=﹣4﹣（﹣6）=﹣4+6=2． 故选C

点评：此题考查了有理数的混合运算，有理数的混合运算首先弄清运算顺序，先乘方，再乘

除，最后算加减，有括号先算括号里边的，同级运算从左到右依次计算，然后利用各种运算法则计算，有时可以利用运算律来简化运算．

30、（13年安徽省8分、15）计算：2sin300

+（—1）2

—22-

31、（2013福省福州16）（1）计算：

；

考点：整式的混合运算；实数的运算；零指数幂． 分析：（1）原式第一项利用零指数幂法则计算，第二项利用负数的绝对值等于它的相反数计算，最后一项化为最简二次根式，计算即可得到结果； 解答：解：（1）原式=1+4﹣2=5﹣2； 点评：此题考查了整式的混合运算，以及实数的运算，

32、（2013?衢州）﹣23

÷|﹣2|×（﹣7+5）

33、（2013甘肃兰州21）（1）计算：（﹣1）2013﹣2﹣1+sin30°+（π﹣3.14）0

考点：实数的运算；零指数幂；负整数指数幂；特殊角的三角函数值． 分析：（1）先计算负整数指数幂、零指数幂以及特殊角的三角函数值，然后计算加减法； 解答：解：（1）原式=﹣1﹣++1=0；

34、（2013年佛山市）计算：[

]

)24()2(521

3

-÷----+?．

分析：根据负整数指数幂以及绝对值、乘方运算法则等性质，先算乘方，再算乘除，最后算加法得出即可

解：2×[5+（﹣2）3]﹣（﹣|﹣4|÷2﹣1

=2×（5﹣8）﹣（﹣4÷）=﹣6﹣（﹣8）=2． 点评：此题主要考查了实数运算，本题需注意的知识点是：负整数指数幂时，a ﹣p

=

35、(2013年深圳市)计算：|-8|+1)3

1(--4?45sin -0)20122013(-

解析：

36、(2013年广东湛江)()2

1--．. 解：原式631=+-

8=

37、（2013?南宁）计算：20130

﹣+2cos60°+（﹣2）

+2×﹣3

38、（2013?六盘水）（1）+（2013﹣π）0

﹣﹣×

=3

39、（2013?黔东南州）（1）计算：sin30°﹣2﹣1+（﹣1）0+；

40、（2013?常德）计算；（π﹣2）0++（﹣1）2013﹣（）﹣2．

41、（2013?张家界）计算：．

×﹣

42、（2013?株洲）计算：．

×

43、（2013?苏州）计算：（﹣1）3+（+1）0+．

+1，

+1

44、（2013?宁夏）计算：．

．

45、（2013?滨州）（计算时不能使用计算器）

计算：．

+2，

+2

．

46、（2013菏泽）（1）计算：

考点：实数的运算；零指数幂；负整数指数幂；特殊角的三角函数值．

分析：（1）求出每部分的值，再代入求出即可；

解答：解：（1）原式=﹣3×+1+2+

=2+；

点评：本题考查了二次根式的性质，零整数指数幂，负整数指数幂，特殊角的三角函数值．47、（2013?巴中）计算：．

48、（2013?遂宁）计算：|﹣3|+．

×

49、（2013?温州）（1）计算：+（）+（）0

=2﹣；

50、（2013?广安）计算：（）﹣1+|1﹣|﹣﹣2sin60°．

﹣×

51、（2013?泸州）计算：．

52、（2013?眉山）计算：2cos45°﹣+（﹣）﹣1+（π﹣3.14）0．

×

﹣4+1=53、（2013?自贡）计算：

= 1 ．

﹣×﹣（﹣﹣2+54、（2013?内江）计算：

．

﹣1+

55、(2013年黄石)011tan 30(2013)()3

π----+

解析：原式3213=+--+ ·

··························································· （5分） 4= （2分）

56、（2013凉山州）计算：

．

考点：实数的运算；零指数幂；特殊角的三角函数值． 专题：计算题．

分析：原式第一项表示2平方的相反数，第二项利用特殊角的三角函数值化简，第三项先计算绝对值里边的式子，再利用绝对值的代数意义化简，第四项利用零指数幂法则计算，即可得到结果．

解答：解：原式=﹣4﹣

+3+1+

=0．

点评：此题考查了实数的运算，涉及的知识有：零指数、负指数幂，平方根的定义，绝对值

的代数意义，熟练掌握运算法则是解本题的关键．

57、（2013四川南充，15，6分）计算（－1）2013+（2sin30°+21）－38+（3

1

）1- 解析：解：原式=－1+1－2+3 ……………4′ =1 ……………6′

(2013浙江丽水)计算：0

)2

1

(28-+--

58、（2013?曲靖）计算：2﹣1

+|﹣|++（）0

．

+59、（2013?昆明）计算：﹣2sin30°．

×=2

60、（2013济宁）计算：（2﹣）

2012

（2+）

2013

﹣2﹣（）0

．

考点：二次根式的混合运算；零指数幂．

分析：根据零指数幂、绝对值、整数指数幂、二次根式的混合运算，分别进行计算，再把所得的结果合并即可． 解答：解：（2﹣

）

2012

（2+）

2013

﹣2﹣（）0

=[（2﹣

）（2+）]

2012

（2+）﹣﹣1 =2+﹣﹣1 =1．

点评：此题考查了二次根式的混合运算，用到的知识点是零指数幂、绝对值、整数指数幂、

二次根式的混合运算，关键是熟练掌握有关知识和公式．

61、（1-4实数的比较与运算·2013东营中考）

计算：(

)1

02 3.142sin 603π-??

?+-- ???

分析：（1）123

21

3()3

2

-==

，0

( 3.14)1π-=

，sin 60?=

=.

（1）解： 原式

=

(3+1212---

=

3

+112

-+ =

3

2

…………………………3分 点拨：（1）分别根据零指数幂、负整数指数幂、特殊角的三角函数值计算出各数，再根据实数混合运算的法则进行解答即可.

62、（2013山西，19(1)，5

1453??

?- ???

.

【解析】解：原式

1- =1-1=0

63、（2013达州）计算：

2

1tan 603-??

-?+ ???

解析：原式＝1＋9＝10

64、（绵阳市2013年）（1）计算：)

2

1

2

12sin 45

-?

-+-

?+；

解： 原式= - 122 +|1- 1

2

2 |×2( 2 +1)

= - 1

4 +( 2 -1) ×2( 2 +1)

= - 1

4 +2[（ 2 ）2 -12]

= 2- 1

4

= 74

65、（德阳市2013年）计算：一12013＋（12

）一2

一｜33tan60° 解析：

高考英语真题分类汇编各种题型全汇总

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