2011 AMC 10A 第 16 题

Question

下列哪一项等于

\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}

？

Which of the following is equal to

\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}?

3\sqrt2

2\sqrt6

\dfrac{7\sqrt2}{2}

3\sqrt3

6

将整个表达式平方。

Square the whole expression

乘积

(9-6\sqrt2)(9+6\sqrt2)

很小。

The product

(9-6\sqrt2)(9+6\sqrt2)

is small

由于式子中含有平方根，可尝试把每个根号内的式子写成完全平方。

注意原式可改写为

\begin{aligned} & \sqrt{6 - 6\sqrt2 + 3} \\ &{}+ \sqrt{6 + 6\sqrt2 + 3} \end{aligned}\text{。}

因式分解并化简得

\sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2}

= \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6\text{。}

所以正确答案是 B。

Since we have square roots, we can try to change the inside of each radical to be a perfect square.

Note that we can rewrite the expression as

\begin{aligned} & \sqrt{6 - 6\sqrt2 + 3} \\ &{}+ \sqrt{6 + 6\sqrt2 + 3}. \end{aligned}

Factoring and simplifying gives us

\sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2}

= \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6.

Thus, B is the correct answer.

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