Question

Даны множества: А={3; 5; 6; 8; 12; 15}, В={3; 6; 8; 14, 20} и се {3;5;6;8; 17; 19). Найдите: 1) (A\B) (A/C);2)(A/B) (A/C) ​ Смотреть ответ vladkonoval17 vladkonoval17 Відповідь: аічікві Покрокове пояснення: ап

Answer 1

1) \( (A\B) \cup (A/C) \)

To find \( (A\B) \cup (A/C) \), we first need to calculate \( A\B \) and \( A/C \).

1. \( A\B \) (the set of elements that are in A but not in B): - \( A = \{3, 5, 6, 8, 12, 15\} \) - \( B = \{3, 6, 8, 14, 20\} \)

Therefore, \( A\B = \{5, 12, 15\} \).

2. \( A/C \) (the set of elements that are in A but not in C): - \( C = \{3, 5, 6, 8, 17, 19\} \)

Therefore, \( A/C = \{12, 15\} \).

Now, we can find \( (A\B) \cup (A/C) \): - \( (A\B) \cup (A/C) = \{5, 12, 15\} \cup \{12, 15\} = \{5, 12, 15\} \).

So, \( (A\B) \cup (A/C) = \{5, 12, 15\} \).

2) \( (A/B) \cup (A/C) \)

To find \( (A/B) \cup (A/C) \), we first need to calculate \( A/B \) and \( A/C \).

1. \( A/B \) (the set of elements that are in A but not in B): - \( A = \{3, 5, 6, 8, 12, 15\} \) - \( B = \{3, 6, 8, 14, 20\} \)

Therefore, \( A/B = \{5, 12, 15\} \).

2. \( A/C \) (the set of elements that are in A but not in C): - \( C = \{3, 5, 6, 8, 17, 19\} \)

Therefore, \( A/C = \{12, 15\} \).

Now, we can find \( (A/B) \cup (A/C) \): - \( (A/B) \cup (A/C) = \{5, 12, 15\} \cup \{12, 15\} = \{5, 12, 15\} \).

So, \( (A/B) \cup (A/C) = \{5, 12, 15\} \).