Question

Exercitiul 28 urgent

Answer 1

Răspuns:

A∪B= {1,2,3,4,5,6,7,8}

A\(A∩B) = {1,2} adica multimea A are in plus (fata de ce au in comun A si B ) elementele {1,2}  

=> B nu are  {1,2}

B\(A∩B) = {7, 8} adica multimea B are aceste elemente {7, 8} (iar intersectia multimilor nu )

=> A nu are {7,8}

=> A= {1,2,3,4,5,6}

B={3,4,5,6,7,8}

Explicație pas cu pas:

Answer 2

$\it A\cup B=\{1,2,3,4,5,6,7,8\} \Rightarrow \begin{cases}\it A\subset \{1,2,3,4,5,6,7,8\} \\ \\\it B\subset \{1,2,3,4,5,6,7,8\}\end{cases}\ \ \ \ (1) \\ \\ \\ A\backslash (A\cap B)=\{1,2\} \Rightarrow \begin{cases}\it \{1,2\}\subset A\\ \\\it \{1,2\}\not\subset B\end{cases}\ \ \ \ (2) \\ \\ \\ B\backslash (A\cap B)=\{7,8\} \Rightarrow \begin{cases}\it \{7,8\}\subset B\\ \\\it \{7,8\}\not\subset A\end{cases}\ \ \ \ (3)$

Din  relațiile  (1), (2), (3) ⇒ A = {1, 2, 3, 4, 5, 6}  și B= {3, 4, 5, 6, 7, 8}.