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〖1/1算式〗
作业:求算式 2*500/sqrt(3)-320-320*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)-320-320*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)
=2*500/sqrt3-320-320*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)
=2*500/sqrt3-320-320*0.05^2/0.0075*((0.1^2/0.05^2)+1)
=2*500/sqrt3-320-320*0.05^2/0.0075*(4+1)
=2*500/sqrt3-320-320*0.05^2/0.0075*5
=2*500/1.732051-320-320*0.05^2/0.0075*5
=2*500/1.732051-320-320*0.0025/0.0075*5
=1000/1.732051-320-320*0.0025/0.0075*5
=577.350205-320-320*0.0025/0.0075*5
=577.350205-320-0.8/0.0075*5
=577.350205-320-106.666667*5
=577.350205-320-533.333335
=257.350205-533.333335
=-275.98313 答案:2*500/sqrt(3)-320-320*0.05^2/(0.1^2-0.05^2)*((0.1^2/0.05^2)+1)=-275.98313你的问题在这里没有得到解决?请到 热门难题 里面看看吧!