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☆算术计算1题
〖1/1算式〗
作业:求算式 2/sqrt(3)*500*ln1-320+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/sqrt(3)*500*ln1-320+320*0.05^2/(0.1^2-0.05^2)*(0.1^2/0.05^2-1)
=2/sqrt3*500*ln1-320+320*0.05^2/(0.1^2-0.05^2)*(0.1^2/0.05^2-1)
=2/sqrt3*500*ln1-320+320*0.05^2/0.0075*(0.1^2/0.05^2-1)
=2/sqrt3*500*ln1-320+320*0.05^2/0.0075*3
=2/1.732051*500*ln1-320+320*0.05^2/0.0075*3
=2/1.732051*500*0-320+320*0.05^2/0.0075*3
=2/1.732051*500*0-320+320*0.0025/0.0075*3
=1.1547*500*0-320+320*0.0025/0.0075*3
=577.35*0-320+320*0.0025/0.0075*3
=0-320+320*0.0025/0.0075*3
=0-320+0.8/0.0075*3
=0-320+106.666667*3
=0-320+320.000001
=-320+320.000001
=0.000001 答案:2/sqrt(3)*500*ln1-320+320*0.05^2/(0.1^2-0.05^2)*(0.1^2/0.05^2-1)=0.000001你的问题在这里没有得到解决?请到 热门难题 里面看看吧!