AMC10 2021 B

AMC10 2021 B · Q2

AMC10 2021 B · Q2. It mainly tests Absolute value, Exponents & radicals.

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

求 $\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}$ 的值。

(A) 0 0

(B) 4\sqrt{3}-6 4\sqrt{3}-6

(D) 4\sqrt{3} 4\sqrt{3}

(E) 4\sqrt{3}+6 4\sqrt{3}+6

Answer

Correct choice: (D)

正确答案：（D）

Solution

Note that the square root of any number squared is always the absolute value of the squared number because the square root function will only return a nonnegative number. By squaring both $3$ and $2\sqrt{3}$, we see that $2\sqrt{3}>3$, thus $3-2\sqrt{3}$ is negative, so we must take the absolute value of $3-2\sqrt{3}$, which is just $2\sqrt{3}-3$. Knowing this, the first term in the expression equals $2\sqrt{3}-3$ and the second term is $3+2\sqrt3$, and summing the two gives $\boxed{\textbf{(D)} ~4\sqrt{3}}$.

注意，任何数的平方根总是该数的绝对值，因为平方根函数只返回非负数。通过平方 $3$ 和 $2\sqrt{3}$，我们看到 $2\sqrt{3}>3$，因此 $3-2\sqrt{3}$ 是负数，所以我们必须取 $3-2\sqrt{3}$ 的绝对值，即 $2\sqrt{3}-3$。知道这一点，表达式第一项等于 $2\sqrt{3}-3$，第二项是 $3+2\sqrt{3}$，两者相加得到 $\boxed{\textbf{(D)} ~4\sqrt{3}}$。

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