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作业:求算式 2*500/sqrt(3)-350-350*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)-350-350*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)
=2*500/sqrt3-350-350*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)
=2*500/sqrt3-350-350*0.05^2/0.0075*((0.1^2/0.05^2)+1)
=2*500/sqrt3-350-350*0.05^2/0.0075*(4+1)
=2*500/sqrt3-350-350*0.05^2/0.0075*5
=2*500/1.732051-350-350*0.05^2/0.0075*5
=2*500/1.732051-350-350*0.0025/0.0075*5
=1000/1.732051-350-350*0.0025/0.0075*5
=577.350205-350-350*0.0025/0.0075*5
=577.350205-350-0.875/0.0075*5
=577.350205-350-116.666667*5
=577.350205-350-583.333335
=227.350205-583.333335
=-355.98313 答案:2*500/sqrt(3)-350-350*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)=-355.98313你的问题在这里没有得到解决?请到 热门难题 里面看看吧!