ข้อสอบคณิต O-NET - 2560

$$\begin{gathered} \mathrm{แก้อสมการเงื่อนไขใน} C \mathrm{จะได้} \\ x = -1, 4 \mathrm{แต่เอกภพสัมพัทธ์} \\ \mathrm{คือ} \mathrm{จำนวนนับ} \to \mathrm{เหลือ} C = \left\{4\right\} \end{gathered}$$

$$\begin{gathered} 1. A\cap C = \left\{1, 2, 3, \dots , 10\right\} \cap \left\{4\right\} = \left\{4\right\} \\ B\cap C = \left\{4, 8, 12, 16, 20\right\} \cap \left\{4\right\} = \left\{4\right\} \to \mathrm{ถูก} \\ 2. B\cup C = \left\{4, 8, 12, 16, 20\right\} \cap \left\{4\right\} \\ = \left\{4, 8, 12, 16, 20\right\} = B \to \mathrm{ถูก} \\ 3. A\cap B = \left\{1, 2, 3, \dots , 10\right\} \cap \left\{4, 8, 12, 16, 20\right\} \\ = \left\{4, 8\right\} \to \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} 4. B-A = \left\{4, 8, 12, 16, 20\right\}-\left\{1, 2, 3, \dots , 10\right\} \\ = \left\{12, 16, 20\right\} \to \mathrm{ถูก} \\ 5. \left(A\cap C\right)\cup B = \left[\left(\left\{1, 2, 3, \dots , 10\right\}\right) \cap \left\{4\right\}\right] \cup \left\{4, 8, 12, 16, 20\right\} \\ = \left\{4\right\} \cup \left\{4, 8, 12, 16, 20\right\} \\ = \left\{4, 8, 12, 16, 20\right\} \to \mathrm{ผิด} \end{gathered}$$