Question

下式的值是多少？ $\begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2} \end{aligned}\text{？}$

What is the value of $\begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2}? \end{aligned}$

$0$

$4\sqrt{3}-6$

$6$

$4\sqrt{3}$

$4\sqrt{3}+6$

记住 $\sqrt{u^2}=|u|$ 。

Remember that $\sqrt{u^2}=|u|$

分别判断 $3-2\sqrt3$ 和 $3+2\sqrt3$ 的正负。

Decide the signs of $3-2\sqrt3$ and $3+2\sqrt3$ separately

因为 $\sqrt{u^2}=|u|$ ，所以原式等于 $|3-2\sqrt3|+|3+2\sqrt3|\text{。}$ 又因为 $2\sqrt3>3$ ，所以它化为 $(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3\text{。}$

所以正确答案是 D。

Because $\sqrt{u^2}=|u|,$ the expression equals $|3-2\sqrt3|+|3+2\sqrt3|.$ Since $2\sqrt3>3,$ this becomes $(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3.$

Thus, the correct answer is D .