AMC12 2010 A

AMC12 2010 A · Q4

AMC12 2010 A · Q4. It mainly tests Absolute value, Exponents & radicals.

If $x<0$, then which of the following must be positive?

若 $x<0$，则下列哪个一定为正？

(A) $\frac{x}{|x|}$ $\frac{x}{|x|}$

(B) $-x^{2}$ $-x^{2}$

(D) $-x^{-1}$ $-x^{-1}$

(E) $\sqrt[3]{x}$ $\sqrt[3]{x}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

$x$ is negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use $-1$. $\textbf{(A)} \Rightarrow \frac{-1}{|-1|} = -1$ $\textbf{(B)} \Rightarrow -(-1)^2 = -1$ $\textbf{(C)} \Rightarrow -2^{(-1)} = -\frac{1}{2}$ $\textbf{(D)} \Rightarrow -(-1)^{(-1)} = 1$ $\textbf{(E)} \Rightarrow \sqrt[3]{-1} = -1$ Obviously only $\boxed{\textbf{(D)}}$ is positive.

$x$ 为负，因此我们可以把一个负值代入每个表达式，找出为正的那个。设取 $-1$。 $\textbf{(A)} \Rightarrow \frac{-1}{|-1|} = -1$ $\textbf{(B)} \Rightarrow -(-1)^2 = -1$ $\textbf{(C)} \Rightarrow -2^{(-1)} = -\frac{1}{2}$ $\textbf{(D)} \Rightarrow -(-1)^{(-1)} = 1$ $\textbf{(E)} \Rightarrow \sqrt[3]{-1} = -1$ 显然只有 $\boxed{\textbf{(D)}}$ 为正。

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