2013 AMC 12A 第 8 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

8.

已知 xxyy 是不同的非零实数,且 x+2x=y+2yx + \dfrac{2}{x} = y + \dfrac{2}{y}, 求 xyxy

Given that xx and yy are distinct nonzero real numbers such that x+2x=y+2y,x + \dfrac{2}{x} = y + \dfrac{2}{y}, what is xy?xy?

14\dfrac{1}{4}

12\dfrac{1}{2}

11

22

44

答案:D
知识点:代数变形因式分解
难度评级:1400
解答:

同乘 xyxyx2y+2y=xy2+2xx^2 y + 2y = xy^2 + 2x, 所以 x2yxy22x+2y=(xy)(xy2)=0. \begin{gathered} x^2 y - xy^2 - 2x + 2y \\ = (x - y)(xy - 2) \\ = 0. \end{gathered}

因为 xyx \ne y, 所以 xy=2xy = 2

因此,正确答案是 D

Multiplying by xyxy gives x2y+2y=xy2+2x,x^2 y + 2y = xy^2 + 2x, so x2yxy22x+2y=(xy)(xy2)=0. \begin{gathered} x^2 y - xy^2 - 2x + 2y \\ = (x - y)(xy - 2) \\ = 0. \end{gathered}

Since xy,x \ne y, it follows that xy=2.xy = 2.

Thus, the correct answer is D.

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