1950 AMC 12 第 10 题

Question

将

\dfrac{\sqrt3-\sqrt2}{\sqrt3}

的分子有理化后，最简形式的分母为：

After rationalizing the numerator of

\dfrac{\sqrt3-\sqrt2}{\sqrt3},

the denominator in simplest form is:

\sqrt3(\sqrt3+\sqrt2)

\sqrt3(\sqrt3-\sqrt2)

3-\sqrt3\sqrt2

3+\sqrt6

以上答案均不正确

None of these answers

分子、分母同乘原分子的共轭式

Multiply the numerator and denominator by the conjugate of the numerator

分子变为

(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=1

The numerator becomes

(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=1

分子、分母同乘原分子的共轭式，得到

\begin{aligned} &\frac{\sqrt3-\sqrt2}{\sqrt3} \cdot\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ &=\frac{1}{\sqrt3(\sqrt3+\sqrt2)}\\ &=\frac{1}{3+\sqrt6} \end{aligned}\text{。}

因此，正确答案是 D。

Multiplying by the conjugate of the numerator gives

\begin{aligned} &\frac{\sqrt3-\sqrt2}{\sqrt3} \cdot\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ &=\frac{1}{\sqrt3(\sqrt3+\sqrt2)}\\ &=\frac{1}{3+\sqrt6}. \end{aligned}

Thus, the correct answer is D.

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