1990 AIME 第 4 题

Question

求下列方程的正数解：

\begin{aligned}&\frac1{x^2-10x-29}\\&\quad+\frac1{x^2-10x-45}\\&\quad-\frac2{x^2-10x-69}=0\end{aligned}\text{。}

Find the positive solution to

\begin{aligned}&\frac1{x^2-10x-29}\\&\quad+\frac1{x^2-10x-45}\\&\quad-\frac2{x^2-10x-69}=0.\end{aligned}

答案

令

y=x^2-10x

，使三个分母只相差常数

Set

y=x^2-10x

so the three denominators differ only by constants

先合并前两个分数，再消去分母

Combine the first two fractions before clearing denominators

令

y=x^2-10x

。合并前两个分数并消去非零分母，得到

(y-37)(y-69)

与

(y-29)(y-45)

相等。展开并约去

y^2

，得到

y=39

。因此

x^2-10x-39=0

，即

(x-13)(x+3)=0

。正数解为

13

。

Set

y=x^2-10x.

Combining the first two fractions and clearing the nonzero denominators gives an equality between

(y-37)(y-69)

and

(y-29)(y-45).

Expanding and canceling

y^2

yields

y=39.

Thus

x^2-10x-39=0,

so

(x-13)(x+3)=0.

The positive solution is

13.

其他年份的第 4 题