猿辅导2017年春季初联训练营作业题解答-7: "高次方程组"

 

1、解方程组 $$\begin{cases}xy + x + y = 1\\ yz + y + z = 5\\ zx + z + x = 2 \end{cases}$$

解答：$$\begin{cases}(x+1)(y+1) = 2\\ (y+1)(z+1) = 6\\ (z+1)(x+1) = 3 \end{cases}$$ $$\Rightarrow (x+1)^2(y+1)^2(z+1)^2 = 36$$ $$\Rightarrow (x+1)(y+1)(z+1) = \pm6$$ $$\Rightarrow \begin{cases}x+1 = \pm1\\ y+1 = \pm2\\ z+1 = \pm3 \end{cases}$$ $$\Rightarrow \begin{cases}x = 0,\ -2\\ y = 1,\ -3\\ z = 2,\ -4 \end{cases}$$ $$\Rightarrow (x, y, z) = (0, 1, 2),\ (-2, -3, -4).$$

 

 

2、解方程组 $$\begin{cases}y = \displaystyle{4x^2\over 1+4x^2}\\ z = \displaystyle{4y^2 \over 1+4y^2} \\ x = \displaystyle{4z^2 \over 1+4z^2}\end{cases}$$

解答：

当$x = y = z = 0$时成立。下面讨论不为零的情形：$$\begin{cases}\dfrac{1+4x^2}{4x^2} = \dfrac{1}{y}\\ \dfrac{1+4y^2}{4y^2} = \dfrac{1}{z}\\ \dfrac{1+4z^2}{4z^2} = \dfrac{1}{x} \end{cases}$$ $$\Rightarrow \begin{cases}1 + \dfrac{1}{4x^2} = \dfrac{1}{y}\\ 1 + \dfrac{1}{4y^2} = \dfrac{1}{z}\\ 1 + \dfrac{1}{4z^2} = \dfrac{1}{x} \end{cases}$$ $$\Rightarrow \frac{1}{4x^2} + \frac{1}{4y^2} + \frac{1}{4z^2} + 3 = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$ $$\Rightarrow \left(\frac{1}{2x} - 1\right)^2 + \left(\frac{1}{2y} - 1\right)^2 + \left(\frac{1}{2z} - 1\right)^2 = 0$$ $$\Rightarrow x= y = z = \frac{1}{2}.$$ 综上，$(x, y, z) = (0, 0, 0)$, $\left(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right)$.

 

 

3、解方程组 $$\begin{cases}x + y + z = 9\\ x^2 + y^2 + z^2 = 41\\ x^3 + y^3 + z^3 = 189 \end{cases}$$

解答：$$xy + yz + zx = \frac{1}{2}\cdot\left[(x+y+z)^2 - (x^2 + y^2 + z^2)\right] = 20.$$ $$\Rightarrow x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$ $$\Rightarrow 189 - 3xyz = 9\cdot(41 - 20) = 189. \Rightarrow xyz = 0.$$ 由对称性，不妨设$z = 0$, $$\begin{cases}x+ y = 9\\ xy = 20 \end{cases} \Rightarrow \begin{cases}x_1 = 4\\ y_1 = 5 \end{cases},\ \begin{cases}x_2 = 5\\ y_2 = 4 \end{cases}.$$ 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 $$\begin{cases}x + 2017^{x} = y + 2017^{y}\\ x^2 + xy + y^2 = 12 \end{cases}$$

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.