Question

方程的各根倒数之和是多少？ $\dfrac{2003}{2004}x + 1 + \dfrac{1}{x} = 0?$

What is the sum of the reciprocals of the roots of the equation $\dfrac{2003}{2004}x + 1 + \dfrac{1}{x} = 0?$

$-\dfrac{2004}{2003}$

$-1$

$\dfrac{2003}{2004}$

$1$

$\dfrac{2004}{2003}$

设 $a = \dfrac{2003}{2004}$ 。两边乘以 $x$ ，得 $ax^2 + x + 1 = 0$ 。

若根为 $r$ 和 $s$ ，则由韦达定理 $r + s = -\dfrac{1}{a}$ ，且 $rs = \dfrac{1}{a}$ 。

倒数和为 $\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-1/a}{1/a} = -1$ 。

所以正确答案是 B。

Let $a = \dfrac{2003}{2004}.$ Multiplying the equation by $x$ gives $ax^2 + x + 1 = 0.$

If the roots are $r$ and $s,$ then by Vieta's formulas $r + s = -\dfrac{1}{a}$ and $rs = \dfrac{1}{a}.$

The sum of reciprocals is $\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-1/a}{1/a} = -1.$

Thus, the correct answer is B.