100道初一数学计算题
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(1)()2
2--= (2)3
112⎛⎫
⎪⎝⎭
-=
(3)()9
1- = (4)()4
2-- =
(5)()
2003
1-= (6)()2
332-+-=
(7)()3
3131-⨯--= (8)()2
233-÷- =
(9))2()3(32-⨯-= (10)22)2
1
(3-÷-=
(11)()()33
2
2222+-+--
(12)235(4)0.25(5)(4)8⎛⎫-⨯--⨯-⨯- ⎪⎝⎭
(13)()3
4255414-÷-⎪⎭
⎫ ⎝⎛-÷
(14)()⎪⎭
⎫ ⎝⎛-÷----72132224
6
(15)()()()3
3
2
20132-⨯+-÷--- (16) []
24)3(26
1
1--⨯-
-
(17)])3(2[)]215.01(1[2--⨯⨯-- (18)
(19)()()()3
3
2
20132-⨯+-÷--- (20)22)2(3---;
(21)]2)33()4[()10(222⨯+--+-; (22)])2(2[3
1
)5.01()1(24--⨯⨯---;
(23)9
4
)211(42415.0322⨯-----+-; (24)20022003)2()2(-+-;
(25))2()3(]2)4[(3)2(223-÷--+-⨯--; (26)200420094)25.0(⨯-.
(27)()025242313
2.⨯--÷-⎛⎝ ⎫⎭⎪+⎡⎣⎢⎢⎤⎦
⎥⎥ (28)()()----⨯-221410222
332222()(3)(3)
33
÷--+-
(29) ()()()-⨯÷-+-⎛⎝ ⎫
⎭
⎪⨯-÷-312031331223
2
325..
(30) ()()()-⎛⎝ ⎫
⎭
⎪⨯-⨯-⨯-212052832.
(31)
(32)(56)(79)---
(33)(3)(9)(8)(5)-⨯---⨯- (34)35
15()26
÷-+
(35)5231591736342
--+- (36)()()2
2431)4(2-+-⨯---
(37)4
1
1)8()54()4()125.0(25⨯-⨯-⨯-⨯-⨯
(38)如果0)2(12=-++b a ,求20112010()-3ab a b a a ++-()的值
33
1
82(4)8
-÷--
(39)已知|1|a +与|4|b -互为相反数,求b a 的值。
(40)2234.0)2.1()211(922÷---⨯ (41)12
11
1110|11101211|-+-
(42)5]36)65121197(45[÷⨯+-- (43) )4
1
()35(12575)125(72-⋅-+⨯--⨯
(44))3
2()87()12787431(-+-÷-- (45)41
31211-+-
(46)()1-⎪⎭
⎫ ⎝⎛-÷2131 (47) 22128(2)2⎛⎫
-⨯-+÷- ⎪⎝⎭
(48)15
64358
-÷⨯ (49))4955.5(1416.34955.61416.3-⨯+⨯
(50)100()()222---÷3)2(32-+⎪⎭
⎫
⎝⎛-÷
(51)113
(5)77(7)12()3322
-⨯+⨯--÷-
(52)2012201313
(2)(0.5)(6)714
-⨯-+-⨯
(53)32201211
1()()(1)(2)(1)22
16⎡⎤--÷--⨯-÷-⎢⎥⎣⎦
(54)2
2
2121(3)242433⎛⎫⎛⎫
-÷⨯-+-⨯- ⎪ ⎪⎝⎭⎝⎭
(55))12()4332125(-⨯-+
(56)(20)(3)(5)(7)-++---+ (57)3712
()()14263
-+----
(58)1
( 6.5)(2)()(5)3
-⨯-÷-÷-
(59)若7a =,3b =,求a +b 的值.
(60)已知│a +1│与│b -2│互为相反数,求a -b 的值.
(61) (-12)÷4×(-6)÷2; (62)235(4)0.25(5)(4)8⎛⎫-⨯--⨯-⨯- ⎪⎝⎭
(63)111311123124244⎛⎫⎛⎫⎛⎫⎛⎫--+----- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭
(64)
2
22121(3)242433⎛⎫⎛⎫
-÷⨯-+-⨯- ⎪ ⎪⎝⎭⎝⎭
(65)999×374-53
÷(0.5)3
(66)已知|m+5|+(n-3)2
=0,m 2x+y
n 与n x m 3
是同类项,则x
2017
+y
2018
-(m+n )的值
(67)532)2(1---+-+;
(68)(-5)×(-7)-5×(-6)
(69)()25.05832-÷⎪⎭
⎫
⎝⎛-÷⎪⎭⎫ ⎝⎛-
(70)()⎪⎭
⎫
⎝⎛----+⎪⎭⎫ ⎝⎛-⋅-21221232
.
(71)2
2
2121(3)242433⎛⎫⎛⎫
-÷⨯-+-⨯- ⎪ ⎪⎝⎭⎝⎭