Question

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

Answer 1

x^2+2yz=x ; (1)
y^2+2zx=z ; (2)
z^2+2xy=y (3)
(1)+(2)+(3):
(x+y+z)^2=(x+y+z)
所以x+y+z=1或0
I>
当x+y+z=1时,x=1-y-z,代入(2),(3)
y^2+z=2z^2+2yz (4)
z^2+y=2y^2+2yz (5)
(4)-(5):
(y-z)(y+z)-(y-z)=2(z-y)(y+z)
即：(y-z)(3y+3z-1)=0
y=z或y+z=1/3
当y+z=1/3时,x=2/3,代入(1)：yz=1/9,无解,舍去.
当y=z时,(2)式化为：y^2+2xy=y,当y=z=0时,x=1,满足原方程组.
当y+2x=1时,因为x+2y=1,故解为x=y=z=1/3,同样满足方程组.
II>
当x+y+z=0时,代入(2),(3)得：
y^2=z+2z^2+2yz (4)
z^2=y+2y^2+2yz (5)
(4)-(5),整理：(y-z)(3y+3z-1)=0
当y=z时,(2)式化为y^2+2xy=y,当y=z=0时,x=0,满足原方程组.
当y+2x=1时,因为x+2y=0,故解为x=2/3,y=z=-1/3,同样满足方程组.
当y+z=1/3时,x=-1/3,yz=-2/9,所以y=-1/3,z=2/3或y=2/3,z=-1/3,均舍去.
综上所述,原方程组解为.
x=1,y=z=0;
x=y=z=1/3;
x=y=z=0
x=2/3,y=z=-1/3
共四组