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二一二01/16×5又3/4×7又2015/2016×5.757÷(1/57)能...

[3/4-(1/4+3/8)]÷7/16=(6/8-2/8-3/8)*16/7=1/8*16

12*5/11 = 60/11 7.6*(2*1/4)+ 7*3/5 -7*3/4= 7.6/2 + 7*(3/5-3/4)= 3.8 + 7*(-3/20)= 19/5 - 21/20=(19*4 -21 )/20= 55/20= 11/4 =2又3/4(0.75÷3-0.2)*(1-4/5)= (0.25-0.2)* (1/5) 注:

(2又2016分之1*3.6+3又5分之3*1又2016分之2015)÷4分之3=3.6*2015/1008*4/3=14/9

=7/16*8/3+1/7*3/37-15/4*14/5*21/11=7/6+3/259-21/11=-12493/17094

2又5/6-(2/3+7/12)=2又5/6-15/12=2又10/12-15/12=1又7/12,15又1/4-5又1/2+7又5/6=(15-5+7)+(1/4+5/6-1/2)=17+7/12=17又7/12

题目:[5又2分之1*(Δ+2又3/14)÷1又4/7]÷(14-8又1/4-4.75)=13求△的值解答:(5.5*(△+2又3/14)/1又4/7)÷(148又1/4一4.75)=13(5.5*(△+2又3/14)/(1又4/7)÷(1413)=13(5.5*(△+2又3/14)/(1又4/7)=13则5.5*(△+2又3/14)=13*7/115.5*(△+2又3/14)=91/11则△+31/14=182/61△=182/6131/14△=657/854

3.75*4又2/5+1.6*3又3/4=3.75*4.4+1.6*3.75=3.75*(4.4+1.6)=3.75*6=22.51又3/7+2又4/15+4又4/7+3又2/15=(1又3/7+4又4/7)+(2又4/15+3又2/15)=6+27/5=6+5.4=11.4

分子=1*2*3*4*(1+2+4)=1*2*3*4*7分母=1*3*5*7*(1+2+4)=1*3*5*7*7所以,原式=1*2*3*4*7/1*3*5*7*7=8/35

39/5÷[8/3*13/8+2x-11/10]-2/15÷2/3=1.839/5÷[13/3+16x/3-1.1]-0.2=1.839/5÷[97/33+16x/3]=2[97/33+16x/3]=39/1016x/3=317/330x=317/1760

1. 41又2/3÷5/3+71又3/4÷7/4+91又4/5÷9/5 =(40+5/3)*3/5+(70+7/4)*4/7+(90+9/5)*5/9=24+1+40+1+50+1=1172. 1/5*27+3/5*41 =3/5*9+3/5*41=3/5*(9+41)=3/5*50=303. 5/11*1/20+3/11*1/2+5/11*1/2=1/11*(5/20+3/2+5/2)=1/11*(1/4+4)=1/11*17/4

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