ข้อสอบคณิต 2 - มีนาคม 2563

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

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

$$\begin{gathered} {\left(\frac{1}{2}+\sqrt{3}\right)}^2 = {\left(\frac{1}{2}\right)}^2+2\left(\frac{1}{2}\right)\left(\sqrt{3}\right)+{\left(\sqrt{3}\right)}^2 \\ = \frac{1}{4}+\sqrt{3}+3 = \frac{13}{4}+\sqrt{3} \\ \mathrm{จะได้} \left(2-\sqrt{3}\right)+\left(\frac{13}{4}+\sqrt{3}\right) = \frac{21}{4} \to \mathrm{ตอบ} 3 \end{gathered}$$