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

$3\sqrt2$

$2\sqrt6$

$\dfrac{7\sqrt2}{2}$

$3\sqrt3$

$6$

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

因式分解并化简得 $\sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2}$ $= \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6.$

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.