Question

解方程组 x^2+2yz=x ,y^2+2zx=z ,z^2+2xy=y

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
此时解为：(1,0.0),(1/3.1/3,1/3)
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