Question

下列比值的值是多少：

$\dfrac{2^{2001} \cdot 3^{2003}}{6^{2002}}\text{？}$

What is the value of the ratio

$\dfrac{2^{2001} \cdot 3^{2003}}{6^{2002}}?$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{2}$

知识点：指数

将 $6^{2002}$ 写成 $2^{2002}\cdot 3^{2002}$ 。

Write $6^{2002}$ as $2^{2002}\cdot 3^{2002}$

约去分子和分母中相同的 $2$ 和 $3$ 的幂。

Cancel the common powers of $2$ and $3$ from top and bottom

因为 $6^{2002} = 2^{2002}\cdot 3^{2002}$ ，所以 $\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}\text{。}$

所以正确答案是 E。

Since $6^{2002} = 2^{2002}\cdot 3^{2002},$ the ratio becomes $\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}.$

Thus, the correct answer is E.