AMC12 2000 A

AMC12 2000 A · Q11

AMC12 2000 A · Q11. It mainly tests Linear equations, Rational expressions.

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$?

两个非零实数 $a$ 和 $b,$ 满足 $ab = a - b$。以下哪一项可能是 $\frac {a}{b} + \frac {b}{a} - ab$ 的值？

(A) $-2$ $-2$

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

(D) $\frac{1}{2}$ $\frac{1}{2}$

(E) 2 2

Answer

Correct choice: (E)

正确答案：（E）

Solution

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}$. Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}$. 另一种方法是对 $b$ 解方程，得到 $b = \frac{a}{a+1};$ 然后将其代入表达式并化简，答案为 $2.$

Topics