Question

x+y+zx=1/2 y+z+xy=1/2 z+x=yz=1/2

Answer 1

解方程组: x+y+zx = 1/2 ①, y+z+xy = 1/2 ②, z+x+yz = 1/2 ③.
顺次由①-②, ②-③, ③-①得: z-x = -x(y-z), x-y = -y(z-x), y-z = -z(x-y).
∴z-x = -x(y-z) = xz(x-y) = -xyz(z-x), 即(z-x)(1+xyz) = 0,
∴z-x = 0或1+xyz = 0.

1) 若z-x = 0, 有x-y = -y(z-x) = 0, 故x = y = z.
代回①得2x+x² = 1/2, 解得x = -1±√6/2.
可以验证x = y = z = -1±√6/2是方程组的两组解.

2) 若xyz+1 = 0. 设x+y+z = p, xy+yz+zx = q.
①+②+③得2p+q = 2(x+y+z)+(xy+yz+zx) = 3/2 ④.
y·①+z·②+x·③得(xy+yz+zx)+(x²+y²+z²)+3xyz = (x+y+z)/2,
即p²-q-3 = (x+y+z)²-(xy+yz+zx)+3xyz = p/2.
将④代入得p²-(3/2-2p)-3 = p/2, 即0 = 2p²+3p-9 = (2p-3)(p+3).

假设p+3 = 0, 由①有z(x-1) = 1/2-(x+y+z) = 7/2, 同理x(y-1) = 7/2, y(z-1) = 7/2.
相乘得q+5 = -xyz+(xy+yz+zx)-(x+y+z)+1 = xyz(x-1)(y-1)(z-1) = 343/8.
q = 303/8, 2p+q ≠ 3/2, 与④矛盾.
因此2p-3 = 0, p = 3/2, 代回④解得q = -3/2.

由根与系数关系, 方程组x+y+z = p = 3/2, xy+yz+zx = q = -3/2, xyz = -1解
恰好是一元三次方程2x³-3x²-3x+2 = 0的三个根.
因式分解得: 2x³-3x²-3x+2 = (x+1)(x-2)(2x-1), 故三个根为-1, 1/2, 2.
代回原方程组验证可得x = -1, y = 2, z = 1/2及其轮换是原方程的三组解.
(而x = -1, y = 1/2, z = 2及其轮换不是解).

综上, 原方程组共有5组解:
x = y = z = -1+√6/2;
x = y = z = -1-√6/2;
x = -1, y = 2, z = 1/2;
x = 2, y = 1/2, z = -1;
x = 1/2, y = -1, z = 2.