Question

方程 $\begin{aligned} a(b-c)x^2+b(c-a)x\\ {}+c(a-b)&=0 \end{aligned}$ 的两个根为 $1$ 和：

The two roots of the equation $\begin{aligned} a(b-c)x^2+b(c-a)x\\ {}+c(a-b)&=0 \end{aligned}$ are $1$ and:

$\dfrac{b(c-a)}{a(b-c)}$

$\dfrac{a(b-c)}{c(a-b)}$

$\dfrac{a(b-c)}{b(c-a)}$

$\dfrac{c(a-b)}{a(b-c)}$

$\dfrac{c(a-b)}{b(c-a)}$

使用两根之积，不必直接解二次方程

Use the product of the roots rather than solving the quadratic

两根之积等于常数项除以首项系数

The product is the constant coefficient divided by the leading coefficient

由韦达定理，两根之积为 $\frac{c(a-b)}{a(b-c)}\text{。}$ 因为其中一个根是 $1$ ，所以另一个根等于这个乘积。

因此，正确答案是 D。

By Vieta’s formulas, the product of the two roots is $\frac{c(a-b)}{a(b-c)}.$ Since one root is $1,$ the other root equals this product.

Thus, the correct answer is D.