Question

解方程组 x^2+2yz=x , y^2+2zx=z , z^2+2xy=y

求助数学高手帮忙，谢谢了

Answer 1

三式相加：x^2+y^2+z^2+2xy+2xz+2yz=x+y+z
即(x+y+z)^2=(x+y+z)
得(x+y+z)(x+y+z-1)=0
因此x+y+z=0， 或x+y+z=1
2)-3): y^2-z^2+2x(z-y)=z-y, 得：(y-z)(y+z-2x+1)=0, 得y=z或y+z-2x+1=0
因此分四种情况：
1)x+y+z=0, y=z, 得：x=-2y, 代入1) 4y^2+2y^2=-2y, 得：3y^2+y=0, 得;y=0, -1/3
此时猜州解为：(0.0.0). (2/3, -1/3.-1/3)
2)x+y+z=0, y+z-2x+1=0, 两式相减：3x-1=0,得x=1/3, y+z=-1/3, 2yz=x-x^2=1/3-1/9=-2/9
即yz=-1/9, 解得：y, z=(-1+√5)/6, (-1-√5)/6
此时解为：(1/3, (-1+√5)/6, (-1-√5)/6), (1/3, (-1-√5)/6, (-1+√5)/6),
3)x+y+z=1, y=z, 得x=1-2y, 代入2): y^2+2y(1-2y)=y,得：y=0, 1/3
此时祥悔解为：(1,0.0), (1/3.1/穗宴蔽3,1/3)
4)x+y+z=1, y+z-2x+1=0, 两式相减：3x-1=1, 得x=2/3, y+z=1/3, 2yz=x-x^2=2/3-4/9=2/9
即yz=1/9, 此时无实根