三元方程組難題求解

5(x + 1/x) = 12(y + 1/y)
5yx² + 5y = 12xy² + 12x
xy (5x - 12y) = 12x - 5y ......(1)
同理 ,
yz (12y - 13z) = 13y - 12z ...(2)
由 xy + yz + zx = 1 ...☆ 得 :
xy = 1 - z(x+y) .........(3)
yz = 1 - x(y+z) .........(4)
;
代 (3) 入 (1) :
[1 - z(x+y)] (5x - 12y) = 12x - 5y
5x - 12y - z(x+y) (5x - 12y) = 12x - 5y
- z (x+y) (5x - 12y) = 7x + 7y
- z (5x - 12y) = 7
5zx - 12yz = - 7 .......(5)
代 (4) 入 (2) 同理得 :
- x (12y - 13z) = 1
12xy - 13zx = - 1 .....(6);
(5)*13 + (6)*5 :
- 156yz + 60xy = - 96
13yz - 5xy = 8 .........(7)
☆* 5 + (7) :
18yz + 5zx = 13 ......(8)
(8) - (5) :
30yz = 20
yz = 2/3 , 代入(7) 及 (8) :
xy = 2/15 及
zx = 1/5
故
yz * zy * zx = (2/3)(2/15)(1/5)
(xyz)² = 4/225
xyz = ± 2/15
xyz / (yz) = (± 2/15) / (2/3)
x = ± 1/5
xyz / (xz) = (± 2/15) / (1/5)
y = ± 2/3
xyz / (xy) = (± 2/15) / (2/15)
z = ± 1
故 (x，y，z) = (±1/5，±2/3，±1)

(x,y,z)=(1/5,2/3,1) 或 (-1/5,-2/3,-1)

(x，y，z)=(1/5，2/3，1)