Question

Mulțimea A={x aparține numerelor naturale atunci fractura 3x+12supra 21 este subunitara}

Answer 1

$A = \Big\{x \in \mathbb_{N} \Big| \dfrac{3x+12}{21} \rightarrow fractie subunitara\Big\} \\ \\ 3x+12\ \textless \ 21 \Rightarrow 3x\ \textless \ 21-12 \Rightarrow 3x\ \textless \ 9 \Rightarrow x \ \textless \ \dfrac{9}{3} \Rightarrow x\ \textless \ 3 \\ \\ \Rightarrow \boxed{A = \Big\{0,1,2\Big\}}$

Answer 2

[tex]\it \dfrac{3x+12}{21} \ \textless \ 1,\ \ x =? \\\;\\ \\\;\\ x=0 \Longrightarrow \dfrac{3\cdot0+12}{21} \ \textless \ 1 \Longrightarrow \dfrac{12}{21} \ \textless \ 1 \ \ \ (A) \\\;\\ \\\;\\ x=1 \Longrightarrow \dfrac{3\cdot1+12}{21} \ \textless \ 1 \Longrightarrow \dfrac{15}{21} \ \textless \ 1 \ \ \ (A) \\\;\\ \\\;\\ x=2 \Longrightarrow \dfrac{3\cdot2+12}{21} \ \textless \ 1 \Longrightarrow \dfrac{18}{21} \ \textless \ 1 \ \ \ (A) \\\;\\ \\\;\\ x=3 \Longrightarrow \dfrac{3\cdot3+12}{21} \ \textless \ 1 \Longrightarrow \dfrac{21}{21} \ \textless \ 1 \ \ \ (F) [/tex]

A = {0,  1,  2}