解方程组2X+YX^2=Y 2Y+ZY^2=Z 2Z+XZ^2=X
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/01 13:13:23
解方程组2X+YX^2=Y 2Y+ZY^2=Z 2Z+XZ^2=X
这个题手算太困难了.我用MATLAB算出答案后发现实数解只有1组:
x=0,y=0,z=0
其余8组解全是虚数解,共9组解.
因为解析解太长,好几百位,故我给出数值
需要解析解的话百度HI我.这里给你一个x的虚数解析解给你看看.
x=
- (15*(7 - (224 + (224*27^(1/2)*i)/27)^(1/3)/2 - 56/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) + (3^(1/2)*(112/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) - ((224*27^(1/2)*i)/27 + 224)^(1/3))*i)/2)^(1/2))/4 + 5*(7 - (224 + (224*27^(1/2)*i)/27)^(1/3)/2 - 56/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) + (3^(1/2)*(112/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) - ((224*27^(1/2)*i)/27 + 224)^(1/3))*i)/2)^(3/2) - (7 - (224 + (224*27^(1/2)*i)/27)^(1/3)/2 - 56/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) + (3^(1/2)*(112/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) - ((224*27^(1/2)*i)/27 + 224)^(1/3))*i)/2)^(5/2)/4
x=0,y=0,z=0
其余8组解全是虚数解,共9组解.
因为解析解太长,好几百位,故我给出数值
需要解析解的话百度HI我.这里给你一个x的虚数解析解给你看看.
x=
- (15*(7 - (224 + (224*27^(1/2)*i)/27)^(1/3)/2 - 56/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) + (3^(1/2)*(112/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) - ((224*27^(1/2)*i)/27 + 224)^(1/3))*i)/2)^(1/2))/4 + 5*(7 - (224 + (224*27^(1/2)*i)/27)^(1/3)/2 - 56/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) + (3^(1/2)*(112/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) - ((224*27^(1/2)*i)/27 + 224)^(1/3))*i)/2)^(3/2) - (7 - (224 + (224*27^(1/2)*i)/27)^(1/3)/2 - 56/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) + (3^(1/2)*(112/(3*(224 + (224*27^(1/2)*i)/27)^(1/3)) - ((224*27^(1/2)*i)/27 + 224)^(1/3))*i)/2)^(5/2)/4
解方程组2X+YX^2=Y 2Y+ZY^2=Z 2Z+XZ^2=X
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