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(完整)100道初一数学计算题

(1)22= (2)3112-=

(3)91 = (4)42 =

(5)20031= (6)2332=

(7)33131= (8)2233 =

(9))2()3(32= (10)22)21(3=

(11)3322222

(12)235(4)0.25(5)(4)8

(13)34255414

(14)721322246

(15)33220132 (16) 24)3(2611

(17)])3(2[)]215.01(1[2 (18)

(19)33220132 (20)22)2(3;

(21)]2)33()4[()10(222; (22)])2(2[31)5.01()1(24;

(23)94)211(42415.0322; (24)20022003)2()2(;

(25))2()3(]2)4[(3)2(223; (26)200420094)25.0(.

(27)0252423132. (28)221410222

332222()(3)(3)33(29) 3120313312232325..

(30) 212052832.

(31) (32)(56)(79)

(33)(3)(9)(8)(5) (34)3515()26

(35)5231591736342 (36)22431)4(2

(37)411)8()54()4()125.0(25

(38)如果0)2(12ba,求20112010()-3ababaa()的值

33182(4)8(39)已知|1|a与|4|b互为相反数,求ba的值。

(40)2234.0)2.1()211(922 (41)12111110|11101211|

(42)5]36)65121197(45[ (43) )41()35(12575)125(72

(44))32()87()12787431( (45)4131211

(46)12131 (47) 22128(2)2

(48)1564358 (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)322012111()()(1)(2)(1)2216

(54)222121(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)222121(3)242433

(65)999×374-53÷(0.5)3

(66)已知|m+5|+(n-3)2=0,m2x+yn与nxm3是同类项,则x2017+y2018-(m+n)的值

(67)532)2(1;

(68)(-5)×(-7)-5×(-6)

(69)25.05832

(70)21221232.

(71)222121(3)242433

(72))12()4332125(

(73)235(4)0.25(5)(4)8

(74)111311123124244;

(75)222121(3)242433;

(76)(-5)×(-8)×0×(-10)×(-15);

(77)(-3)×(-4)×(-5)+(-5)×(-7)

(78)(-0.1)×(-1)×(-100)-0.•01×(1000).

(79)214×(-134)×(-23)×(-87);

(80)(-12 + 13-14-15)×(-20);

(81)(-313)×(-0.12)×(-214)×3313;

(82)(79- 56 + 34- 718)×(-36).

(83)-56×(12-225-0.6)

(84)(+12)×|-23|×214×(-513);

(85)(-118)×3(-23)×(-113)

(86))8(12)11(9 (87)(-213)×(-37)

(88)0×(-13.52)= (89)(-1)2020×a =

(90)(-3.25)×(+213)

(91)(-185.8)×(-3645)×0×(-25)

(92))25()7()4( (93) )34(8)53(

(94))1514348(43 (95))8(45)201(

(96)(-37)×0.125×(-213)×(-8)

(97)53)8()92()4()52(8

(98)(-0.25)×0.5×(-427)×4

(99)(-4)×(-18.36)×2.5

(100)(-29)×(-18)+(-511)×(-3)×215

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