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〖1/1算式〗
作业:求算式 (40*0.0179-0.35^2)*(1-3245.022/(6490.044+4004))^(4/3)*(2*1619.073/1800-1) 的值.
题型:数学计算
解:
(40*0.0179-0.35^2)*(1-3245.022/(6490.044+4004))^(4/3)*(2*1619.073/1800-1)
=0.5935*(1-3245.022/(6490.044+4004))^(4/3)*(2*1619.073/1800-1)
=0.5935*(1-3245.022/10494.044)^(4/3)*(2*1619.073/1800-1)
=0.5935*0.690775^(4/3)*(2*1619.073/1800-1)
=0.5935*0.690775^1.333333*(2*1619.073/1800-1)
=0.5935*0.690775^1.333333*0.79897
=0.5935*0.610635715614176*0.79897
=0.362412*0.79897
=0.289556 答案:(40*0.0179-0.35^2)*(1-3245.022/(6490.044+4004))^(4/3)*(2*1619.073/1800-1)=0.289556你的问题在这里没有得到解决?请到 热门难题 里面看看吧!