Question

Exercițiul 6 va rog foarte frumos dau corona

Answer 1

$A=\{x \in N/ \frac{6}{x} \in N \}\\\\ \frac{6}{x} \in N \Longrightarrow \ \ x \in D_6 \to x \in \{ 1;2;3;6\} \\x \in N\\\\Pentru \ x=1 \ \to \frac{6}{1} =6 \in N \\\\ x=2 \to \frac{6}{2}=3 \in N \\\\ x=3\to \frac{6}{3}=2 \in N \\\\ x=6 \to \frac{6}{6}=1 \in N \\\\ \boxed{A=\{1;2;3;6 \}}$


$B=\{x \in N/\frac{12}{2x+1} \in N\} \\\\\\ \frac{12}{2x+1} \in N \ \Longrightarrow 2x+1 \in D_{12} \to 2x+1 \in \{1;2;3;4;6;12\} \\ x\in N \\\\ Pentru \ 2x+1=1 \ \textless \ =\ \textgreater \ 2x=0 \to x =0 \in N \\\\ 2x+1=2\ \textless \ =\ \textgreater \ 2x=1 \to x=\frac{1}{2} \not \in N \\\\ 2x+1=3\ \textless \ =\ \textgreater \ 2x=2 \to x=1 \in N \\\\ 2x+1=4 \ \textless \ =\ \textgreater \ 2x=3\to x=\frac{3}{2} \not \in N \\\\ 2x+1=6\ \textless \ =\ \textgreater \ 2x=5 \to x=\frac{5}{2} \not \in N \\\\ 2x+1=12\ \textless \ =\ \textgreater \ 2x=11 \to x=\frac{11}{2} \not \in N \\\\\\ \boxed{B=\{0;1\}}$


$C=\{x \in N/ \ \overline{3x1} \ \in N\} \\\\ \overline{3x1} \in N \ \ \Longrightarrow x \in \{0;1;2;3;4;5;6;7;8;9\} \\ x\in N \\\\\\ \boxed{C=\{301;311;321;331;341;351;361;371;381;391\}}$


$B-(A \cap C)=? \\\\\\ A\cap C= \varnothing \\\\\\ B- \varnothing= B \\\\\\ \boxed{B-(A \cap C)= \{0;1\}}$