จงหา \( A \cup B \) เมื่อกำหนคเซต \( A \) และเซต \( B \) ต่อไปนี้ 1. \( \mathbf{A}=\{1,2\} \quad \mathrm{B}=\{3,4\} \) \( A \cup B= \) \( \qquad \) 2. \( \mathbf{A}=\{1,3\} \quad \mathrm{B}=\{1,3,5\} \) \( A \cup B= \) \( \qquad \) 3. \( A=\left\{x \in I^{+} \mid x \leq 4\right\} \quad B=\left\{x \mid x \in I^{+}\right\} \) \( A \cup B= \) \( \qquad \) 4. \( A=\{x \in I \mid-5 \leq x \leq 5\} \quad B=\{x \in I \mid x>2\} \) \( A \cup B= \) \( \qquad \) 5. \( A=\{x \in I \mid x>2\} \quad B=\{x \in I \mid x<5\} \) \( A \cup B= \) \( \qquad \) 6. \( A=\left\{x \in I \mid 15 x^{2}-2 x-1=0\right\} \quad B=\left\{x \in I \mid x^{2}-9 x+14=0\right\} \) \( A \cup B= \) \( \qquad \) 7. \( A=\{0,1,3,5\} \quad B=\{1,3,4,6\} \) \( A \cup B= \) \( \qquad \)

Alright, I need to find the union of sets A and B for each of the given problems. Let's tackle them one by one. **Problem 1:** - **Sets:** \( A = \{1, 2\} \), \( B = \{3, 4\} \) - **Union:** \( A \cup B \) Since there are no common elements between A and B, the union will simply be the combination of all elements from both sets. - **Result:** \( \{1, 2, 3, 4\} \) **Problem 2:** - **Sets:** \( A = \{1, 3\} \), \( B = \{1, 3, 5\} \) - **Union:** \( A \cup B \) Here, both sets share elements 1 and 3. The union will include all unique elements from both sets. - **Result:** \( \{1, 3, 5\} \) **Problem 3:** - **Sets:** \( A = \{x \in \mathbb{I}^+ \mid x \leq 4\} \), \( B = \{x \mid x \in \mathbb{I}^+\} \) First, let's interpret the notation: - \( \mathbb{I}^+ \) represents the set of positive integers. - \( A = \{x \in \mathbb{I}^+ \mid x \leq 4\} \) means all positive integers less than or equal to 4. - \( B = \{x \mid x \in \mathbb{I}^+\} \) means all positive integers. So, \( A = \{1, 2, 3, 4\} \) and \( B = \mathbb{I}^+ \). - **Union:** \( A \cup B = \mathbb{I}^+ \) Since \( B \) already includes all positive integers, the union is just \( B \). - **Result:** \( \mathbb{I}^+ \) **Problem 4:** - **Sets:** \( A = \{x \in \mathbb{I} \mid -5 \leq x \leq 5\} \), \( B = \{x \in \mathbb{I} \mid x > 2\} \) First, let's determine the elements of each set: - \( A = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} \) - \( B = \{3, 4, 5, 6, 7, \ldots\} \) Now, the union \( A \cup B \) will include all elements from both sets without duplication. - **Result:** \( \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, \ldots\} \) **Problem 5:** - **Sets:** \( A = \{x \in \mathbb{I} \mid x > 2\} \), \( B = \{x \in \mathbb{I} \mid x < 5\} \) Let's identify the elements: - \( A = \{3, 4, 5, 6, 7, \ldots\} \) - \( B = \{1, 2, 3, 4\} \) The union \( A \cup B \) will include all elements from both sets. - **Result:** \( \{1, 2, 3, 4, 5, 6, 7, \ldots\} \) **Problem 6:** - **Sets:** \( A = \{x \in \mathbb{I} \mid 15x^2 - 2x - 1 = 0\} \), \( B = \{x \in \mathbb{I} \mid x^2 - 9x + 14 = 0\} \) First, solve the equations to find the elements of each set. For \( A \): - \( 15x^2 - 2x - 1 = 0 \) Using the quadratic formula: - \( x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 15 \cdot (-1)}}{2 \cdot 15} = \frac{2 \pm \sqrt{4 + 60}}{30} = \frac{2 \pm \sqrt{64}}{30} = \frac{2 \pm 8}{30} \) So, \( x = \frac{10}{30} = \frac{1}{3} \) or \( x = \frac{-6}{30} = -\frac{1}{5} \) Since \( x \) must be an integer (\( \mathbb{I} \)), there are no integer solutions for \( A \). - **Set A:** \( \emptyset \) For \( B \): - \( x^2 - 9x + 14 = 0 \) Using the quadratic formula: - \( x = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2} = \frac{9 \pm 5}{2} \) So, \( x = \frac{14}{2} = 7 \) or \( x = \frac{4}{2} = 2 \) - **Set B:** \( \{2, 7\} \) Now, the union \( A \cup B \) will include all elements from both sets. - **Result:** \( \{2, 7\} \) **Problem 7:** - **Sets:** \( A = \{0, 1, 3, 5\} \), \( B = \{1, 3, 4, 6\} \) - **Union:** \( A \cup B \) Identify common elements and combine the sets. - **Result:** \( \{0, 1, 3, 4, 5, 6\} \)