ข้อสอบคณิต O-NET - 2554

$$\begin{gathered} \mathrm{จากโจทย์} x^2-4xy+y^2 \\ \mathrm{จัดรูป} = x^2+2xy+y^2-2xy-4xy \\ = {\left(x+y\right)}^2-6xy \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} x = \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}} , y = \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} \\ \mathrm{จะได้} = {\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)}^2-6\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\left(\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right) \end{gathered}$$

$$\begin{gathered} = \left[\frac{{\left(\sqrt{2}+\sqrt{3}\right)}^2+{\left(\sqrt{2}-\sqrt{3}\right)}^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\right]-6 \\ = {\left[\frac{\left(2+2\sqrt{6}+3\right)+\left(2-2\sqrt{6}+3\right)}{2-3}\right]}^2-6 \\ = {\left(\frac{10}{-1}\right)}^2-6 = 94 \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 94 \end{gathered}$$