1983 AMC 12 第 21 题

Question

从下列各数中找出最小的正数

Find the smallest positive number from the numbers below

10-3\sqrt{11}

3\sqrt{11}-10

18-5\sqrt{13}

51-10\sqrt{26}

10\sqrt{26}-51

比较平方，判断所列各差中哪些为正

Compare the squares to determine which listed differences are positive

利用

a-b=\frac{a^2-b^2}{a+b}

将每个正的差有理化

Rationalize each positive difference using

a-b=\frac{a^2-b^2}{a+b}

由于

100\gt99

，选项 A 为正；由于

324\lt325

，选项 C 为负；又因

2601\gt2600

，选项 D 为正而 E 为负。两个正数满足

\begin{aligned} 10-3\sqrt{11} &=\frac1{10+3\sqrt{11}},\\ 51-10\sqrt{26} &=\frac1{51+10\sqrt{26}} \end{aligned}\text{。}

第二个分母较大，所以选项 D 是较小的正数。

所以正确答案是 D。

Since

100\gt99,

choice A is positive; since

324\lt325,

choice C is negative; and since

2601\gt2600,

choice D is positive while E is negative. The two positive values satisfy

\begin{aligned} 10-3\sqrt{11} &=\frac1{10+3\sqrt{11}},\\ 51-10\sqrt{26} &=\frac1{51+10\sqrt{26}}. \end{aligned}

The second denominator is larger, so choice D is the smaller positive number.

Therefore, the correct answer is D.

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