AMC12 2013 A

AMC12 2013 A · Q8

AMC12 2013 A · Q8. It mainly tests Manipulating equations.

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

已知 $x$ 和 $y$ 是不同的非零实数，且 $x + \frac{2}{x} = y + \frac{2}{y}$，求 $xy$？

(A) \frac{1}{4} \frac{1}{4}

(B) \frac{1}{2} \frac{1}{2}

(D) 2 2

(E) 4 4

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Multiplying the given equation by $xy\neq 0$ yields $x^2y+2y=xy^2+2x$. Thus $$ x^2y-2x-xy^2+2y=x(xy-2)-y(xy-2)=(x-y)(xy-2)=0. $$ Because $x-y\neq 0$, it follows that $xy=2$.

答案（D）：将所给方程两边乘以 $xy\neq 0$，得 $x^2y+2y=xy^2+2x$。因此 $$ x^2y-2x-xy^2+2y=x(xy-2)-y(xy-2)=(x-y)(xy-2)=0。 $$ 因为 $x-y\neq 0$，所以 $xy=2$。

Topics

AL.016 Manipulating equations