Question

Determinati elementele urmatoarelor multimi:

b)B={x ∈ ℕ | 8 pe x-2 ∈ ℕ}

c)C={x ∈ ℕ| 3x -2 pe x +4  ∈ ℕ}

d)D={x ∈ ℕ | 4x + 3 pe 5x - 1 ∈ ℕ}

Answer 1

[tex]a)x-2\in D_8=\{1,2,4,8\}\\ x-2=1=>x=3\\ x-2=2=>x=4\\ x-2=4=>x=6\\ x-2=8=>x=10\\ B=\{3,4,6,10\}[/tex]
[tex]b) \frac{3x-2}{x+4} \in N=> \frac{3x+3\cdot 4-3 \cdot 4-2}{x+4} \in N=>\\ \frac{3(x+ 4)-14}{x+4} \in N=> \frac{3(x+4)}{x+4} - \frac{14}{x+4} \in N=>\\ 3- \frac{14}{x+4} \in N=>\frac{14}{x+4} \in N=>x+4\in D_{14}=\{1,2,7,14\}\\ x+4=1=>x=-3\notin N\\ x+4=2=>x=-2 \notin\n\\ x+4=7=>x=3 \in N\\ x+4=14=>x=10\in \n\\ C=\{3,10\} [/tex]
[tex]c) \frac{4x+3}{5x-1}\in N=>5 \cdot \frac{4x+3}{5x-1}\in N=>\frac{20x+15}{5x-1}\in N=>\\ \frac{4 \cdot 5x-4 \cdot 1+4 \cdot 1+15}{5x-1}\in N=> \frac{4(5x-1)+19}{5x-1}\in N=>\\ \frac{4(5x-1)}{5x-1} + \frac{19}{5x-1}\in N => 4 + \frac{19}{5x-1}\in N =>\\ \frac{19}{5x-1}\in N=>5x-1\in D_{19}=\{1,19\}\\ 5x-1=1=>x= \frac{2}{5} \notin N\\ 5x-1=19=>x=4\in N\\ D=\{4\}[/tex]