2002 AMC 12A 第 13 题

Question

两个不同的正数

a

和

b

各自与其倒数相差

1

。

a + b

是多少？

Two different positive numbers

a

and

b

each differ from their reciprocals by

1.

What is

a + b?

1

2

\sqrt{5}

\sqrt{6}

3

答案：C

知识点：分式方程二次方程

难度评级：1500

小提示：

满足

\left|x - \dfrac1x\right| = 1

的数

x

满足

x^2 - x - 1 = 0

或

x^2 + x - 1 = 0

。

A number

x

with

\left|x - \dfrac1x\right| = 1

satisfies

x^2 - x - 1 = 0

or

x^2 + x - 1 = 0

大提示：

两个正解互为倒数；用求根公式把它们相加。

The two positive solutions are reciprocals of each other; add them using the quadratic formula

解答：

正数

x

与其倒数相差

1

时，

x - \dfrac1x = 1

或

x - \dfrac1x = -1

，即

x^2 - x - 1 = 0

或

x^2 + x - 1 = 0

。

正根为

\dfrac{1+\sqrt5}{2}

和

\dfrac{-1+\sqrt5}{2}

，它们互为倒数。它们的和是

a + b = \sqrt5

。

因此，正确答案是 C。

A positive number

x

differs from its reciprocal by

1

when

x - \dfrac1x = 1

or

x - \dfrac1x = -1,

i.e.

x^2 - x - 1 = 0

or

x^2 + x - 1 = 0.

The positive roots are

\dfrac{1+\sqrt5}{2}

and

\dfrac{-1+\sqrt5}{2},

which are reciprocals of each other. Their sum is

a + b = \sqrt5.

Thus, the correct answer is C.

2002 AMC 12A 第 13 题

答案：C

难度评级：1500

小提示：

大提示：

解答：

其他年份的第 13 题