求解三元三次方程组的方法x+y+z=0.17177xy+xz+yz=0.00566xyz=0.0000523182

Answer 1

知识前提：一元三次方程根与系数的关系.
假设x1,x2,x3是一元三次方程 x^3+bx^2+cx+d=0...(1)的三个解,则有：
(x-x1)(x-x2)(x-x3)=0,展开得：x^3 + (-x1 - x2 - x3) x^2+ (x1 x2 + x1 x3 + x2 x3)x - x1 x2 x3 =0...(2)
对比(1)(2)两式得：x1+x2+x3=-b,x1x2+x1x3+x2x3=c,x1x2x3=-d
在本题中,x,y,z可以看作是方程：x^3-0.17177x^2+0.00566x-0.0000523182=0.(3)的三个根.
用公式法解得方程(3)的三根是：x1=0.0187115,x2=0.0212059,x3=0.131853
所以x,y,z是x1,x2,x3的任意组合,
x=0.0187115,y=0.0212059,z=0.131853
x=0.0187115,y=0.131853,z=0.0212059
x=0.0212059,y=0.0187115,z=0.131853
x=0.0212059,y=0.131853,z=0.0187115
x=0.131853,y=0.0187115,z=0.0212059
x=0.131853,y=0.0212059,z=0.0187115