Question

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

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

$0$

$4\sqrt{3}-6$

$6$

$4\sqrt{3}$

$4\sqrt{3}+6$

先利用平方根与绝对值的关系：因为 $\sqrt{u^2}=|u|$ ，所以 $2\sqrt3>3$ 。 $|3-2\sqrt3|+|3+2\sqrt3|.$ $(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3.$

因此原式为所以正确答案是 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 .