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作业:求算式 2*500/sqrt(3)-400-400*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1) 的值.
题型:数学计算
解:
2*500/sqrt(3)-400-400*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)
=2*500/sqrt3-400-400*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)
=2*500/sqrt3-400-400*0.05^2/0.0075*((0.1^2/0.05^2)+1)
=2*500/sqrt3-400-400*0.05^2/0.0075*(4+1)
=2*500/sqrt3-400-400*0.05^2/0.0075*5
=2*500/1.732051-400-400*0.05^2/0.0075*5
=2*500/1.732051-400-400*0.0025/0.0075*5
=1000/1.732051-400-400*0.0025/0.0075*5
=577.350205-400-400*0.0025/0.0075*5
=577.350205-400-1/0.0075*5
=577.350205-400-133.333333*5
=577.350205-400-666.666665
=177.350205-666.666665
=-489.31646 答案:2*500/sqrt(3)-400-400*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)=-489.31646你的问题在这里没有得到解决?请到 热门难题 里面看看吧!