ข้อสอบคณิต O-NET - 2554

$$\begin{gathered} \mathrm{จากโจทย์} {\left(\sqrt{3}-1\right)}^{-2} \\ \mathrm{จะได้} {\left(\sqrt{3}-1\right)}^{-2} = \frac{1}{{\left(\sqrt{3}-1\right)}^2} \\ = \frac{1}{3-2\sqrt{3}+1} \\ = \frac{1}{4-2\sqrt{3}} \end{gathered}$$

$$\begin{gathered} = \frac{1}{\left(2\right)\left(2-\sqrt{3}\right)} \\ = \frac{1\left(2+\sqrt{3}\right)}{\left(2\right)\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)} \\ = \frac{2+\sqrt{3}}{\left(2\right)\left(4-3\right)} \\ = \frac{2+\sqrt{3}}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{\mathrm{อตรรกยะ}}{\mathrm{ตรรกยะ}} = \mathrm{อตรรกยะ} \\ = \frac{2+1.732}{2} \\ = 1.866 \\ \mathrm{ดังนั้น} \mathrm{ตอบข้อ} 2 \end{gathered}$$