Question

求的值。 $10\cdot\left(\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{10}\right)^{-1}?$

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

$3$

$8$

$\dfrac{25}{2}$

$\dfrac{170}{3}$

$170$

答案：C

知识点：分数运算顺序

难度评级：770

解答：

先计算括号内的和： $\begin{aligned} &\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{10} \\ &= \dfrac{5}{10} + \dfrac{2}{10} + \dfrac{1}{10} \\&= \dfrac{4}{5}. \end{aligned}$

因此 $\left(\dfrac{4}{5}\right)^{-1} = \dfrac{5}{4}$ ，最后得到 $10 \cdot \dfrac{5}{4} = \dfrac{25}{2}.$

所以正确答案是 C。

We get that $\begin{aligned} &\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{10} \\ &= \dfrac{5}{10} + \dfrac{2}{10} + \dfrac{1}{10} \\&= \dfrac{4}{5}. \end{aligned}$

Then $\left(\dfrac{4}{5}\right)^{-1} = \dfrac{5}{4}$ and then finally, $10 \cdot \dfrac{5}{4} = \dfrac{25}{2}.$

Thus, C is the correct answer.

2014 AMC 10A 第 1 题

答案：C

难度评级：770

解答：

其他年份的第 1 题