1959 AMC 12 第 16 题

Question

表达式

\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12}\text{，}

化简后为：

The expression

\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},

when simplified, is:

\dfrac{(x-1)(x-6)}{(x-3)(x-4)}

\dfrac{x+3}{x-3}

\dfrac{x+1}{x-1}

1

2

将四个二次式完全因式分解

Factor all four quadratics completely

把除法改写为乘以倒数

Replace division by multiplication by the reciprocal

因式分解并乘以倒数得

\begin{aligned} &\frac{(x-1)(x-2)}{(x-2)(x-3)}\\ &\quad{}\cdot \frac{(x-3)(x-4)}{(x-1)(x-4)}=1 \end{aligned}

对原表达式定义域内的每个值都成立。

因此，正确答案是 D。

Factoring and multiplying by the reciprocal gives

\begin{aligned} &\frac{(x-1)(x-2)}{(x-2)(x-3)}\\ &\quad{}\cdot \frac{(x-3)(x-4)}{(x-1)(x-4)}=1 \end{aligned}

for every value in the domain of the original expression.

Therefore, the correct answer is D.

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