[五年级数学3道计算题]1.李明和王达步行同时从A.B两地出发,相向而行,在离A地52米处相遇,到达对方出发点后,两人立即以原来的速度沿原路返回,又在离A地处44米处相遇,求A.B两地的距离是多少米? 解:(52乘以3加...+阅读
1)2.5*32*0.125 =(2.5*4)*(8*0.125) =10*1 =10 (2)3.5-7+6.5 =3.5+6.5-7 =3 (3)1.2*2.5+0.8*2.5 =2.5(1.2+0.8) =25 (4)8.9*1.25-0.9*1.25 =1.25(8.9-0.9) =10 (5)12.5*7.4*0.8 =12.5*0.8*7.4 =74 (6)6.5*9.5+6.5*0.5 =6.5*(9.5+0.5) =6.5*10 =65 (7)0.35*1.6+0.35*3.4 =0.35*(1.6+3.4) =0.35*5 =1.75 (8)6.72-3.28-1.72 =6.75-(3.28+1.72) =6.75-5 =1.75 (9)0.45+6.37+4.55 =0.45+4.55+6.37 =5+6.37 =11.37 (10)28*12.5-12.5*20 =(28-20)*12.5 =8*12.5 =100 (11)23.65-(3.07+3.65) =23.65-3.65-3.07 =20-3.07 =16.93 (12)(4+0.4*0.25)8*7*1.25 =(4+1)*7*10 =5*7*10 =350 (13)1.65*99+1.65 =1.65*(99+1) =1.65*100 =165 (14)27.85-(7.85+3.4) =27.85-7.85+3.4 =20+3.4 =23.4 (15)48*1.25+50*1.25*0.2*8 7.8*9.9+0.78 =7.8*(9.9+0.1) =7.8*10 =78 (16)4.8*46+4.8*54 =(46+54)*4.8 =100*4.8 =480 (17)673-327-173 =673-(327+173) =673-500 =173 (18)4.44*2.5 =1.11*(4*2.5) =1.11*10 =11.1 (19)3.5÷1.4 =(3.5÷0.7)÷(1.4÷0.7) =5÷2 =2.5 (20)1+2+3+4+5......+99+100 =(1+99)+(2+98)+(3+97)+......+(49+51)+50 =(100÷2-1)*100+50 =4900+50 =4950
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