1、解方程组 $$egin{cases}xy + x + y = 1\ yz + y + z = 5\ zx + z + x = 2 end{cases}$$
解答：$$egin{cases}(x+1)(y+1) = 2\ (y+1)(z+1) = 6\ (z+1)(x+1) = 3 end{cases}$$ $$Rightarrow (x+1)^2(y+1)^2(z+1)^2 = 36$$ $$Rightarrow (x+1)(y+1)(z+1) = pm6$$ $$Rightarrow egin{cases}x+1 = pm1\ y+1 = pm2\ z+1 = pm3 end{cases}$$ $$Rightarrow egin{cases}x = 0, -2\ y = 1, -3\ z = 2, -4 end{cases}$$ $$Rightarrow (x, y, z) = (0, 1, 2), (-2, -3, -4).$$
2、解方程组 $$egin{cases}y = displaystyle{4x^2over 1+4x^2}\ z = displaystyle{4y^2 over 1+4y^2} \ x = displaystyle{4z^2 over 1+4z^2}end{cases}$$
解答：
当$x = y = z = 0$时成立。下面讨论不为零的情形：$$egin{cases}dfrac{1+4x^2}{4x^2} = dfrac{1}{y}\ dfrac{1+4y^2}{4y^2} = dfrac{1}{z}\ dfrac{1+4z^2}{4z^2} = dfrac{1}{x} end{cases}$$ $$Rightarrow egin{cases}1 + dfrac{1}{4x^2} = dfrac{1}{y}\ 1 + dfrac{1}{4y^2} = dfrac{1}{z}\ 1 + dfrac{1}{4z^2} = dfrac{1}{x} end{cases}$$ $$Rightarrow frac{1}{4x^2} + frac{1}{4y^2} + frac{1}{4z^2} + 3 = frac{1}{x} + frac{1}{y} + frac{1}{z}$$ $$Rightarrow left(frac{1}{2x} - 1 ight)^2 + left(frac{1}{2y} - 1 ight)^2 + left(frac{1}{2z} - 1 ight)^2 = 0$$ $$Rightarrow x= y = z = frac{1}{2}.$$ 综上，$(x, y, z) = (0, 0, 0)$, $left(dfrac{1}{2}, dfrac{1}{2}, dfrac{1}{2} ight)$.
3、解方程组 $$egin{cases}x + y + z = 9\ x^2 + y^2 + z^2 = 41\ x^3 + y^3 + z^3 = 189 end{cases}$$
解答：$$xy + yz + zx = frac{1}{2}cdotleft[(x+y+z)^2 - (x^2 + y^2 + z^2) ight] = 20.$$ $$Rightarrow x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$ $$Rightarrow 189 - 3xyz = 9cdot(41 - 20) = 189. Rightarrow xyz = 0.$$ 由对称性，不妨设$z = 0$, $$egin{cases}x+ y = 9\ xy = 20 end{cases} Rightarrow egin{cases}x_1 = 4\ y_1 = 5 end{cases}, egin{cases}x_2 = 5\ y_2 = 4 end{cases}.$$ 综上，$(x, y, z) = (4, 5, 0)$, $(5, 4, 0)$, $(0, 4, 5)$, $(0, 5, 4)$, $(4, 0, 5)$, $(5, 0,4)$.
4、解方程组 $$egin{cases}x + 2017^{x} = y + 2017^{y}\ x^2 + xy + y^2 = 12 end{cases}$$
解答：
易知$x = y$ （可分别讨论$x > y$，$x < y$时均矛盾）。$$Rightarrow 3x^2 = 12 Rightarrow x = pm2.$$ 综上，$(x, y) = (2, 2)$, $(-2, -2)$.
主讲教师:
赵胤, 理学硕士(数学) & 教育硕士(数学), 中国数学奥林匹克一级教练员, 高级中学数学教师资格.