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〖1/1算式〗
作业:求算式 (10)^(3.5048-(4.079-0.301*(8-1.55))*(-50*ln0.9/20)^(1/6.8713)) 的值.
题型:数学计算
解:
(10)^(3.5048-(4.079-0.301*(8-1.55))*(-50*ln0.9/20)^(1/6.8713))
=10^(3.5048-(4.079-0.301*(8-1.55))*(-50*ln0.9/20)^(1/6.8713))
=10^(3.5048-(4.079-0.301*6.45)*(-50*ln0.9/20)^(1/6.8713))
=10^(3.5048-2.13755*(-50*ln0.9/20)^(1/6.8713))
=10^(3.5048-2.13755*0.263402^(1/6.8713))
=10^(3.5048-2.13755*0.263402^0.145533)
=10^1.744457
=55.520964244786924 答案:(10)^(3.5048-(4.079-0.301*(8-1.55))*(-50*ln0.9/20)^(1/6.8713))=55.520964244786924你的问题在这里没有得到解决?请到 热门难题 里面看看吧!