无理式的有理化

公式

\[(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})=a - b \]

\[(\sqrt[3]{a} \pm \sqrt[3]{b})(\sqrt[3]{a^2} \mp \sqrt[3]{ab} + \sqrt[3]{b^2})=a \pm b \]

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例题

1. 设 \(\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = a\)，且 \(x \gt y \gt 0\)，求 \(\frac{y\sqrt{a^2 - 4}}{a - \sqrt{a^2 - 4}}\)

解：

\[\begin{align*} 原式 &= \frac{y \sqrt{a^2 - 4}}{a - \sqrt{a^2 - 4}} \\ &= \frac{y \sqrt{a^2 - 4}(a + \sqrt{a^2 - 4})}{(a - \sqrt{a^2 - 4})(a + \sqrt{a^2 - 4})} \\ &= \frac{y \sqrt{a^2 - 4}(a + \sqrt{a^2 - 4})}{4} \end{align*} \]

\[\begin{align*} \because a^2 - 4 &= \left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\right)^2 - 4 \\ &= \frac{x}{y}+\frac{y}{x} - 2 \\ &= \left(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}\right)^2 \end{align*} \]

\[\therefore \sqrt{a^2-4} = \sqrt{\frac{x}{y}} - \sqrt{\frac{y}{x}} \quad (x \gt y \gt 0) \]

\[\begin{align*} \therefore \frac{y \sqrt{a^2 - 4}}{a - \sqrt{a^2 - 4}} &= \frac{y \sqrt{a^2 - 4}(a + \sqrt{a^2 - 4})}{4} \\ &= \frac{y \left[ \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right) \left( \sqrt{\frac{x}{y}} - \sqrt{\frac{y}{x}} \right) + \left( \sqrt{\frac{x}{y}} - \sqrt{\frac{y}{x}} \right)^2 \right]}{4} \\ &= \frac{y \left(\frac{x}{y} - \frac{y}{x} + \frac{x}{y} - 2 + \frac{y}{x} \right)}{4} \\ &= \frac{y \left(2\frac{x}{y} - 2 \right)}{4} \\ &= \frac{x - y}{2} \end{align*} \]

2. 设 \(f(x)=\sqrt{x}+\sqrt{x+1}\)，求 \(\sum_{i=1}^{n}\frac{1}{f(i)}=\frac{1}{f(1)}+\frac{1}{f(2)}+\cdots+\frac{1}{f(n)}\)

解：

\[\begin{align*} \because \frac{1}{f(x)} &=\frac{1}{\sqrt{x}+\sqrt{x+1}} \\ &=\frac{\sqrt{x}-\sqrt{x+1}}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}-\sqrt{x+1})} \\ &=\frac{\sqrt{x}-\sqrt{x+1}}{x-(x+1)} \\ &=\sqrt{x+1}-\sqrt{x} \end{align*} \]

\[\therefore \sum_{i=1}^{n}\frac{1}{f(i)} =(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{n+1}-\sqrt{n}) \]

裂项相消可得：

\[\sum_{i=1}^{n}\frac{1}{f(i)}=\sqrt{n+1}-1 \]

3. 化简 \(\frac{\sqrt[3]{a^2b}-\sqrt[3]{ab^2}}{\sqrt[3]{ax}-\sqrt[3]{bx}}\)

解：

\[\because (\sqrt[3]{ax}-\sqrt[3]{bx})(\sqrt[3]{(ax)^2}+\sqrt[3]{abx^2}+\sqrt[3]{(bx)^2}) = ax - bx = (a-b)x \]

\[又\because (\sqrt[3]{a^2b}-\sqrt[3]{ab^2})(\sqrt[3]{(ax)^2}+\sqrt[3]{abx^2}+\sqrt[3]{(bx)^2}) = (a-b)\sqrt[3]{abx^2} \]

\[\begin{align*} \therefore原式 &=\frac{\sqrt[3]{a^2b}-\sqrt[3]{ab^2}}{\sqrt[3]{ax}-\sqrt[3]{bx}} \frac{\sqrt[3]{(ax)^2}+\sqrt[3]{abx^2}+\sqrt[3]{(bx)^2}}{\sqrt[3]{(ax)^2}+\sqrt[3]{abx^2}+\sqrt[3]{(bx)^2}} \\ &=\frac{(a-b)\sqrt[3]{abx^2}}{(a-b)x} \\ &=\frac{\sqrt[3]{abx^2}}{x} \end{align*} \]