ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{จากโจทย์} 2{\log }_2\left(x+y\right) = x-y \\ {\log }_2\left(x+y\right) = \frac{x-y}{2} \\ x+y = 2^{\left(\frac{x-y}{2}\right)} \to \boxed{1} \\ \mathrm{จากโจทย์} \left(x+y\right)3^{y-x} = \frac{2}{9} \end{gathered}$$
$$\begin{gathered} -\mathrm{แทน} x+y = 2^{\left(\frac{x-y}{2}\right)} \\ \mathrm{จะได้} 2^{\left(\frac{x-y}{2}\right)}{3}^{y-x} = \frac{2}{9} \\ 2^{\left(\frac{x-y}{2}\right)}·3^{-\left(x-y\right)} = \frac{2}{9} \\ \frac{2^{\left(\frac{x-y}{2}\right)}}{3^{x-y}} = \frac{2^1}{3^2} \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \frac{x-y}{2} = 1 \mathrm{และ} x-y = 2 \\ x-y = 2 \\ \mathrm{ดังนั้น} x-y = 2 \to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} x+y = 2^{\left(\frac{x-y}{2}\right)} \\ -\mathrm{แทน} x-y=2 \\ \mathrm{จะได้} x+y = 2^{\left(\frac{\cancel{2}}{\cancel{2}}\right)} \\ x+y = 2 \to \boxed{3} \end{gathered}$$

$$\begin{gathered} -\mathrm{จาก} \boxed{2},\boxed{3} \mathrm{แก้} 2 \mathrm{สมการ} 2 \mathrm{ตัวแปร} \\ \mathrm{จะได้} x=2 , y=0 \\ \mathrm{ดังนั้น} x^2+y^2 = 2^2+0^2 \\ x^2+y^2 = 4 \end{gathered}$$