以下是关于如何相除和化简根式表达式的高一（10年级）练习题，并附带详细的分步解答。
具有相同根指数的根式除法公式如下：
$${\frac{\sqrt[n]{x}}{\sqrt[n]{y}} = \sqrt[n]{\frac{x}{y}}}$$
通过阅读以下示例，了解如何应用商法则来化简表达式。
示例 1：
$$\frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$
示例 2：
$$\begin{aligned} \frac{\sqrt[3]{54x^5}}{\sqrt[3]{2x^2}} &= \sqrt[3]{\frac{54x^5}{2x^2}} \\ &= \sqrt[3]{\frac{54}{2} \cdot \frac{x^5}{x^2}} = \sqrt[3]{27x^{5-2}} \\ &= \sqrt[3]{27} \sqrt[3]{x^3} = 3x \end{aligned}$$
示例 3：
$$\begin{aligned} \frac{\sqrt{x^2 - 1}}{\sqrt{x + 1}} &= \sqrt{\frac{x^2 - 1}{x + 1}} \\ &= \sqrt{\frac{(x - 1)(x + 1)}{x + 1}} = \sqrt{x - 1} \end{aligned}$$
示例 4：
$$\begin{aligned} \frac{\sqrt[3]{24y^5x^2}}{\sqrt[3]{3y^2x^5}} &= \sqrt[3]{\frac{24y^5x^2}{3y^2x^5}} \\ &= \sqrt[3]{\frac{24}{3} \cdot \frac{y^5}{y^2} \cdot \frac{x^2}{x^5}} = \sqrt[3]{8 y^{5-2} x^{2-5}} \\ &= \sqrt[3]{8 y^3 x^{-3}} = \sqrt[3]{8 \frac{y^3}{x^3}} = 2\frac{y}{x} \end{aligned}$$
利用除法公式化简下列表达式。点击“查看解答”来核对你的练习成果。
化简：$$\frac{\sqrt[3]{128}}{\sqrt[3]{2}}$$
$$\begin{aligned} \frac{\sqrt[3]{128}}{\sqrt[3]{2}} &= \sqrt[3]{\frac{128}{2}} = \sqrt[3]{64} \\ &= \sqrt[3]{2^6} = \sqrt[3]{(2^2)^3} = 2^2 = 4 \end{aligned}$$
化简：$$\frac{\sqrt{162x^5}}{\sqrt{2x}}$$
$$\begin{aligned} \frac{\sqrt{162x^5}}{\sqrt{2x}} &= \sqrt{\frac{162x^5}{2x}} \\ &= \sqrt{\frac{162}{2} \cdot \frac{x^5}{x}} = \sqrt{81x^4} = 9x^2 \end{aligned}$$
化简：$$\frac{\sqrt{x+2}}{\sqrt{x^2-4}}$$
$$\begin{aligned} \frac{\sqrt{x+2}}{\sqrt{x^2-4}} &= \sqrt{\frac{x+2}{x^2-4}} \\ &= \sqrt{\frac{x+2}{(x-2)(x+2)}} = \sqrt{\frac{1}{x-2}} = \frac{1}{\sqrt{x-2}} \end{aligned}$$
化简：$$\frac{\sqrt[4]{32z^6y^3}}{\sqrt[4]{2z^2y^{11}}}$$
$$\begin{aligned} \frac{\sqrt[4]{32z^6y^3}}{\sqrt[4]{2z^2y^{11}}} &= \sqrt[4]{\frac{32z^6y^3}{2z^2y^{11}}} \\ &= \sqrt[4]{\frac{32}{2} \cdot \frac{z^6}{z^2} \cdot \frac{y^3}{y^{11}}} = \sqrt[4]{16 \cdot \frac{z^4}{y^8}} = 2 \frac{z}{y^2} \end{aligned}$$