Question

对实数 $a$ 和 $b$ ，定义 $a \diamond b = \sqrt{a^2 + b^2}$ 。下式的值是多少？ $(5 \diamond 12) \diamond ((-12) \diamond (-5))\text{？}$

For real numbers $a$ and $b,$ define $a \diamond b = \sqrt{a^2 + b^2}.$ What is the value of $(5 \diamond 12) \diamond ((-12) \diamond (-5))?$

$0$

$\dfrac{17}{2}$

$13$

$13\sqrt{2}$

$26$

先计算两个内层运算。

Evaluate the two inner operations first

$5\diamond12$ 和 $(-12)\diamond(-5)$ 都等于 $13$ 。

Both $5\diamond12$ and $(-12)\diamond(-5)$ equal $13$

第一个内层表达式为 $5 \diamond 12 = \sqrt{5^2 + 12^2} = 13\text{，}$ 而同理 $(-12) \diamond (-5)$ $= \sqrt{144 + 25} = 13$ 。

因此 $13 \diamond 13 = \sqrt{13^2 + 13^2} = 13\sqrt{2}\text{。}$

所以正确答案是 D。

Each inner expression evaluates to $5 \diamond 12 = \sqrt{5^2 + 12^2} = 13$ and similarly $(-12) \diamond (-5)$ $= \sqrt{144 + 25} = 13.$

Then $13 \diamond 13 = \sqrt{13^2 + 13^2} = 13\sqrt{2}.$

Thus, D is the correct answer.