作业帮求matlab大神用matlab指导下面的题目
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/09 03:28:28
求matlab大神用matlab指导下面的题目
已知 f=1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+2/3*(-3*c*a+b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a,
g= (((d/(2*a) + b^3/(27*a^3) - (b*c)/(6*a^2))^2 + (c/(3*a) - b^2/(9*a^2))^3)^(1/2) - b^3/(27*a^3) - d/(2*a) + (b*c)/(6*a^2))^(1/3) - b/(3*a) - (c/(3*a) - b^2/(9*a^2))/(((d/(2*a) + b^3/(27*a^3) - (b*c)/(6*a^2))^2 + (c/(3*a) - b^2/(9*a^2))^3)^(1/2) - b^3/(27*a^3) - d/(2*a) + (b*c)/(6*a^2))^(1/3)
证明：f=g
已知 f=1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+2/3*(-3*c*a+b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a,
g= (((d/(2*a) + b^3/(27*a^3) - (b*c)/(6*a^2))^2 + (c/(3*a) - b^2/(9*a^2))^3)^(1/2) - b^3/(27*a^3) - d/(2*a) + (b*c)/(6*a^2))^(1/3) - b/(3*a) - (c/(3*a) - b^2/(9*a^2))/(((d/(2*a) + b^3/(27*a^3) - (b*c)/(6*a^2))^2 + (c/(3*a) - b^2/(9*a^2))^3)^(1/2) - b^3/(27*a^3) - d/(2*a) + (b*c)/(6*a^2))^(1/3)
证明：f=g
先定义a b c 然后f -g是否等于0.判断
再问： 用matlab的simple()方法，化简的结果不是0。。。
再答： 没办法，我试过了，matlab的符号计算能力就这样。 simplify: (36*a*b*c - 8*b^3 - 108*a^2*d + 12*3^(1/2)*a*(27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)^(1/2))^(1/3)/(6*a) - (2^(2/3)*(-(27*a^2*d + 2*b^3 - 3*3^(1/2)*a^3*((27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)/a^4)^(1/2) - 9*a*b*c)/a^3)^(1/3))/6 - (2*a*c - (2*b^2)/3)/(a*(36*a*b*c - 8*b^3 - 108*a^2*d + 12*3^(1/2)*a*(27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)^(1/2))^(1/3)) + (2^(1/3)*(3*a*c - b^2))/(3*a^2*(-(27*a^2*d + 2*b^3 - 3*3^(1/2)*a^3*((27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)/a^4)^(1/2) - 9*a*b*c)/a^3)^(1/3)) 都是一串。 要不，你用maple试试？
再问： 用matlab的simple()方法，化简的结果不是0。。。
再答： 没办法，我试过了，matlab的符号计算能力就这样。 simplify: (36*a*b*c - 8*b^3 - 108*a^2*d + 12*3^(1/2)*a*(27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)^(1/2))^(1/3)/(6*a) - (2^(2/3)*(-(27*a^2*d + 2*b^3 - 3*3^(1/2)*a^3*((27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)/a^4)^(1/2) - 9*a*b*c)/a^3)^(1/3))/6 - (2*a*c - (2*b^2)/3)/(a*(36*a*b*c - 8*b^3 - 108*a^2*d + 12*3^(1/2)*a*(27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)^(1/2))^(1/3)) + (2^(1/3)*(3*a*c - b^2))/(3*a^2*(-(27*a^2*d + 2*b^3 - 3*3^(1/2)*a^3*((27*a^2*d^2 - 18*a*b*c*d + 4*a*c^3 + 4*b^3*d - b^2*c^2)/a^4)^(1/2) - 9*a*b*c)/a^3)^(1/3)) 都是一串。 要不，你用maple试试？