Question

se dau multimile A={x apartine N/ 3x+2 supra x-2 apartine Z} B={x apartine N/ 5x+7 supra x-1 apartine N} sa se afle A reunit cu B, A intersectat cu B, A-B si B-a. Va roooogg

Answer 1

Date initiale:

$A= \{ x \in N \;| \; \frac{3x+2}{x-2} \in Z \}$

$B= \{ x \in N \;| \; \frac{5x+7}{x-1} \in N \}$

Determinarea multimii B:

$\frac{3x+2}{x-2} = \frac{3x-6+8}{x-2}=\frac{3x-6}{x-2}+\frac{8}{x-2}=3+\frac{8}{x-2} \in Z \\ < = > (x-2) \vert 8\\ \\= > (x-2) \in \{-8, -4,-2,-1,1,2,4,8\}\\ < = > x \in \{-6,-2,0,1,3,4,6,10\}\\\\dar\;\; x\in N \\= > A= \{ 0,1,3,4,6,10 \}$

Determinarea multimii B:

$\frac{5x+7}{x-1} = \frac{5x-5+12}{x-1}=\frac{5x-5}{x-1}+\frac{12}{x-1}=5+\frac{12}{x-1} \in N \\ < = > (x-1) \vert 12\\ \\= > (x-1) \in \{-12,-6,-4,-3,-2,-1,1,2,3,4,6,12\}\\ < = > x \in \{-11,-5,-3,-2,-1,0,2,3,4,5,7,13\}\\\\dar\;\; x\in N \\= > x \in \{ 0,2,3,4,5,7,13 \}$

Observatie: pt x=0 =>

$5+ \frac{12}{x-1}= 5+\frac{12}{-1}=5-12=-7 \notin N$

Pentru oricare alta valoare gasita conditiile de apartenenta la multimea B sunt indeplinite.

$= > B = \{ 2,3,4,5,7,13 \}$

$A = \{ 0,1,3,4,6,10 \}\\B= \{2,3,4,5,7,13\}$

Raspuns:

$A \cup B= \{ 0,1,2,3,4,5,6,7,10,13 \}\\A \cap B=\{ 3,4\}\\A -B=\{ 0,1,6,10\}\\B-A= \{ 2,5,7,13 \}$