AMC12 2000 A

AMC12 2000 A · Q20

AMC12 2000 A · Q20. It mainly tests Manipulating equations, Algebra misc.

If $x,y,$ and $z$ are positive numbers satisfying \[x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}\] Then what is the value of $xyz$ ?

若$x,y,$和$z$为正数，满足 \[x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}\] 则$xyz$的值是多少？

(A) 2/3 $2/3$

(B) 1 1

(D) 2 2

(E) 7/3 $7/3$

Answer

Correct choice: (B)

正确答案：（B）

Solution

We multiply all given expressions to get: \[(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}\] Adding all the given expressions gives that \[(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}\] We subtract $(2)$ from $(1)$ to get that $xyz + \frac{1}{xyz} = 2$. Hence, by inspection, $\boxed{xyz = 1 \rightarrow B}$. \[\]

将所有给定等式的左边相乘，得到： \[(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}\] 将所有给定等式相加，得到 \[(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}\] 用$(1)$减去$(2)$，得$xyz + \frac{1}{xyz} = 2$。因此可直接看出$\boxed{xyz = 1 \rightarrow B}$。 \[\]

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