[tex]\displaystyle\it\\\\\boxed{\it 2.c}\\\\\frac{3x-1}{2x+1}\in\mathbb{Z}\Leftrightarrow 2x+1|3x-1,~atunci~2x+1~va~divide~si~pe~-2(3x-1)=-6x+2.\\dar,~2x+1|2x+1,~atunci~2x+1~va~divide~si~pe~3(2x+1)=6x+3.\\atunci,~2x+1~va~divide~si~suma~celor~doua~numere,~2x+1|(6x+3-6x+2),\\adica~2x+1|5,~deci~2x+1\in\left\{-5,-1,1,5\right\}\Leftrightarrow 2x\in\left\{-6,-2,0,4\right\}\Leftrightarrow\\x\in\left\{-3,-1,0,2\right\},~deci~A=\left\{-3,-1,0,2\right\}.\\------------------------------\\[/tex]
[tex]\displaystyle\it\\\boxed{\it 3.}\\n=\frac{a}{a+d}+\frac{b}{b+d}+\frac{c}{c+d},~n\in\mathbb{N}.\\a)~se~observa~ca~m+n=\bigg(\frac{a}{a+d}+\frac{b}{b+d}+\frac{c}{c+d}\bigg)+\bigg(\frac{d}{a+d}+\frac{d}{b+d}+\frac{d}{c+d}\bigg),\\adica~m+n=\frac{a+d}{a+d}+\frac{b+d}{b+d}+\frac{c+d}{c+d}=1+1+1=3,~iar~cum\\n\in\mathbb{N},~si~suma~m+n\in\mathbb{N}\implies m\in\mathbb{N}.\\b)~din~a),~pentru~ca~m+n=3,~iar~m~si~n~sunt~numere~naturale,~iar\\3~se~poate~scrie~ca~3+0~sau~2+1,~avem~ca~m~si~n~pot~fi~2~si~1~sau~3~si~0.\\[/tex]
[tex]\displaystyle\it\\avand~ca~numerele~sunt~reale~si~pozitive,~m~sau~n~nu~pot~fi~0.\\deci,~m~si~n~pot~fi~2~si~1.\\atunci,~\boxed{\it|m-n|=1}.\\c)~daca~(a,b,c,d)=(3,1,1,3),~atunci~n~este~un~numar~natural.[/tex]
