Question

下列式子的值是多少？ $10 \cdot \left(\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{10}\right)^{-1}$

What is $10 \cdot \left(\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{10}\right)^{-1}?$

$3$

$8$

$\dfrac{25}{2}$

$\dfrac{170}{3}$

$170$

知识点：分数指数

先把括号内的三个分数相加。

Add the three fractions inside the parentheses first

$\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{8}{10}$ ，然后取倒数并乘以 $10$ 。

$\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{8}{10}$ , then take the reciprocal and multiply by $10$

括号内的和为 $\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{5+2+1}{10}=\dfrac{4}{5}$ 。

它的倒数是 $\dfrac54$ ，所以原式等于 $10\cdot\dfrac54=\dfrac{25}{2}$ 。

所以正确答案是 C。

The sum inside the parentheses is $\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{5+2+1}{10}=\dfrac{4}{5}.$

Its reciprocal is $\dfrac54,$ so the expression equals $10\cdot\dfrac54=\dfrac{25}{2}.$

Thus, the correct answer is C.