Question

化简 $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}\text{。}$

Simplify $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}.$

$\sqrt{x}$

$\sqrt[3]{x^2}$

$\sqrt[27]{x^2}$

$\sqrt[54]{x}$

$\sqrt[81]{x^{80}}$

知识点：根式指数

从内向外，用分数指数重写每个根式。

Rewrite each radical using fractional exponents, working from the inside out

最内层是 $x\sqrt{x} = x^{\frac{3}{2}}$ ，每次开立方都会把指数除以 $3$ 。

The innermost $x\sqrt{x} = x^{\frac{3}{2}},$ and each cube root divides the exponent by $3$

从内向外计算， $x\sqrt{x} = x^{\frac{3}{2}}$ ，它的立方根为 $x^{\frac{1}{2}}$ 。

接着 $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}}$ ，它的立方根仍为 $x^{\frac{1}{2}}$ 。

于是下一层又是 $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}}$ ，开立方后仍为 $x^{\frac{1}{2}} = \sqrt{x}$ 。

所以正确答案是 A。

Working outward, $x\sqrt{x} = x^{\frac{3}{2}},$ and its cube root is $x^{\frac{1}{2}}.$

Then $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},$ whose cube root is again $x^{\frac{1}{2}}.$

Repeating once more, $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},$ whose cube root is $x^{\frac{1}{2}} = \sqrt{x}.$

Thus, the correct answer is A.