[tex]\displaystyle\bf\\Explicatii~1:\\\\x\Big|\Big(x+3\Big)~~care~inseamna: x~divide~pe~\Big(x+3\Big), este ~echivalent~cu\\\\\Big(x+3\Big)~\vdots~x~~care inseamna~\Big(x+3\Big)~este~divizibil~cu~x.\\\\Se~mai~poate~scrie~asa:~~\frac{x+3}{x}\in N~adica~impartirea~este~fara~rest.\\\\Explicatia~2\\\\Metoda~de~rezolvare: \\\\Scriem~fractia~(ca~mai~sus)~pe~care~o~descompunem\\~in~suma~de~2~fractii~din~care~prima~prin~simplificare\\devine~un~numar~natural~si~a~doua~fractie~este~mai~usor~de~rezolvat.\\[/tex]
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[tex]\displaystyle\bf\\Rezolvare:\\a)\\x\Big|\Big(x+3\Big)~~\Longleftrightarrow~~\frac{x+3}{x}\in N\\\\\frac{x+3}{x}=\frac{x}{x}+\frac{3}{x}=1+\frac{3}{x}\\\\\frac{x}{x}=1\in N\\\\\implies \frac{3}{x}\in N\\\\\implies x\in D_3\\\\D_3=\{1;~3\}\\\\\boxed{\bf x\in\{1;~3\}}[/tex]
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[tex]\displaystyle\bf\\b)\\\Big(2x\Big)\Big|\Big(x+5\Big)\\\\\Big(x+5\Big)\vdots\Big(2x\Big)\implies \Big(2(x+5)\Big)\vdots\Big(2x\Big)\\\\\frac{2(x+5)}{2x}\in N\\\\\frac{2x+10}{2x}=\frac{2x}{2x}+\frac{10}{2x}\\\\\frac{2x}{2x}=1\in N~\implies~\frac{10}{2x}\in N\\\\\Big(2x\Big)\in D_{10}\\D_{10}=\{1;~2;~5;~10\}\\\\2x=1~\implies x=\frac{1}{2} \notin N\\\\2x=2~\implies x=\frac{2}{2}=1 \in N\\\\2x=5~\implies x=\frac{5}{2} \notin N\\\\2x=10~\implies x=\frac{10}{2}=5 \in N\\\\\boxed{\bf x\in\{1;~5\}}[/tex]
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[tex]\displaystyle\bf\\\\c)\\\Big(x+2\Big)\Big|\Big(2x+1\Big)\\\\\Big(2x+1\Big)\vdots\Big(x+2\Big)\\\\\frac{2x+1}{x+2}\in N\\\\\frac{2x+1}{x+2}=\frac{2x+4-4+1}{x+2}=\frac{2x+4-3}{x+2}=\\\\=\frac{2x+4}{x+2}-\frac{3}{x+2}=\frac{2(x+2)}{x+2}-\frac{3}{x+2}=\left(2-\frac{3}{x+2}\right)\in N\\\\\Big(x+2\Big)\in D_3\\\\D_3=\{1;~3\}\\\\x+2=1~\implies x=1-1=(-1)\notin N\\\\x+2=3~\implies x=3-1=1\in N\\\\\boxed{\bf x=1}[/tex]
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[tex]\displaystyle\bf\\d)\\\Big(3x+1\Big)\Big|\Big(2x+5\Big)\\\\\Big(2x+5\Big)\vdots\Big(3x+1\Big)\implies \Big(3(2x+5)\Big)\vdots\Big(3x+1\Big)\\\\\frac{3(2x+5)}{3x+1}=\frac{6x+15}{3x+1}=\frac{6x+2+13}{3x+1}=\frac{6x+2}{3x+1}+\frac{13}{3x+1}\\\\\frac{6x+2}{3x+1}=\frac{2(3x+1)}{3x+1}=2\in N\\\\\implies\frac{13}{3x+1}\in N\\\\\Big(3x+1\Big)\in D_{13}\\\\D_{13}=\{1;~13\}\\\\3x+1=1 \implies x=0 \in N\\\\3x+1=13 \implies x=\frac{13-1}{3}=4 \in N\\\\\boxed{\bf x\in \{0;~4\}}[/tex]
