Question

a) 6
_ apartine lui lN iar n Aprtine lui Z
n

b) 12
_ apartine lui lN iar apartine lui Z
n+1

c) 13
_ apartine lui Z iar n apartine lui Z
2n+1

d) 9n+14
_____ apartine lui Z iar n apartine lui Z
3n+1

AJUTATI-MA REPEDE VA ROOOG!

Answer 1

   
[tex]\displaystyle\\ a)\\ \frac{6}{n} \in N \Longrightarrow 6~\vdots~n \Longrightarrow n\in D_6\\\\ D_{6N}=\{1;2;3;6\} \Longrightarrow n \in \{1;2;3;6\}~~\text{(n nu poate fi negativ.)} [/tex]


[tex]\displaystyle\\ b)\\ \frac{12}{n+1}\in N\Longrightarrow 12~\vdots~(n+1)\Longrightarrow (n+1)\in D_{12}\\\\ D_{12N}=\{1;2;3;4;6;12\}\\ n+1=1 \Longrightarrow n=1-1 =\boxed{0}\\ n+1=2 \Longrightarrow n=2-1 =\boxed{1}\\ n+1=3 \Longrightarrow n=3-1 =\boxed{2}\\ n+1=4 \Longrightarrow n=4-1 =\boxed{3}\\ n+1=6 \Longrightarrow n=6-1 =\boxed{5}\\ n+1=12 \Longrightarrow n=12-1 =\boxed{11}\\\\ n\in \{0;1;2;3;5;11\} ~~\text{(n nu poate fi negativ)}[/tex]


[tex]\displaystyle\\ c)\\ \frac{13}{2n+1}\in Z \Longrightarrow 13~\vdots~(2n+1)\Longrightarrow (2n+1)\in D_{13}\\\\ D_{13Z} = \{-13;-1;1;13\}\\ 2n+1=-13\Longrightarrow n= \frac{-13-1}{2}= \frac{-14}{2}= \boxed{-7}\\\\ 2n+1=-1\Longrightarrow n= \frac{-1-1}{2}= \frac{-2}{2}= \boxed{-1}\\\\ 2n+1=1\Longrightarrow n= \frac{1-1}{2}= \frac{0}{2}= \boxed{0}\\\\ 2n+1=13\Longrightarrow n= \frac{13-1}{2}= \frac{12}{2}= \boxed{6}\\\\ n\in\{-7;-1;0;6\}[/tex]


[tex]\displaystyle\\ d)\\ \frac{9n+14}{3n+1} =\frac{9n+3+11}{3n+1}=\\\\ =\frac{9n+3}{3n+1}+\frac{11}{3n+1}=\frac{3(3n+1)}{3n+1}+\frac{11}{3n+1}=3+\frac{11}{3n+1}\\\\ \frac{11}{3n+1}\in Z \Longrightarrow 11~\vdots~(3n+1)\Longrightarrow (3n+1)\in D_{11} [/tex]


[tex]\displaystyle\\ D_{11Z}=\{-11;-1;1;11\}\\\\ 3n+1=-11\Longrightarrow 3n=-11-1\Longrightarrow 3n=-12\Longrightarrow n= \frac{-12}{3}=\boxed{-4}\\ 3n+1=-1\Longrightarrow 3n=-1-1\Longrightarrow 3n=-2\Longrightarrow n= \frac{-2}{3}\notin Z \\ 3n+1=1\Longrightarrow 3n=1-1\Longrightarrow 3n=0 \Longrightarrow n= \frac{0}{3}=\boxed{0}\\ 3n+1=11\Longrightarrow 3n=11-1\Longrightarrow 3n=10 \Longrightarrow n= \frac{10}{3} \notin Z\\ n \in \{-4;0\}[/tex]