Question

若正数 $x$ 、 $y$ 和 $z$ 满足以及则 $xyz$ 等于多少？ $x + \frac{1}{y} = 4,$ $y + \frac{1}{z} = 1,$ $z + \frac{1}{x} = \frac{7}{3},$

If $x,$ $y,$ and $z$ are positive numbers satisfying $x + \frac{1}{y} = 4,$ $y + \frac{1}{z} = 1,$ and $z + \frac{1}{x} = \frac{7}{3},$ then what is $xyz?$

$\dfrac{2}{3}$

$1$

$\dfrac{4}{3}$

$2$

$\dfrac{7}{3}$

答案：B

知识点：方程组代数变形对称性（代数）

难度评级：1970

解答：

三个方程相加得 $\begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}. \end{gathered}$

三个方程相乘得 $4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}.$

展开乘积，得到中间括号中的一组之和就是 $\tfrac{22}{3}$ ，所以 $xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2$ 。 $\begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}. \end{aligned}$

因此 $(xyz - 1)^2 = 0$ ，所以 $xyz = 1$ 。

因此，正确答案是 B。

Adding the three equations gives $\begin{gathered} \left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) \\ {}+ \left(z + \tfrac1x\right) \\ = 4 + 1 + \tfrac73 \\ = \tfrac{22}{3}. \end{gathered}$

Multiplying them gives $4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}.$

Expanding the product, $\begin{aligned} &\left(x + \tfrac1y\right) \\ &\quad {}\cdot \left(y + \tfrac1z\right) \\ &\quad {}\cdot \left(z + \tfrac1x\right) \\ &= xyz \\ &\quad {}+ \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) \\ &\quad {}+ \frac{1}{xyz}. \end{aligned}$ The middle group is the sum $\tfrac{22}{3},$ so $xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2.$

Hence $(xyz - 1)^2 = 0,$ so $xyz = 1.$

Thus, the correct answer is B.

2000 AMC 12 第 20 题

答案：B

难度评级：1970

解答：

其他年份的第 20 题