Question

Îmi puteți explica va rog mult tipurile acestea de exerciții am mâine test și nu le înțeleg deloc nu trebuie sa faceți exercițiul trebuie doar de pilda sa luați punctul a sau care vreți voi și sa îmi explicați și dacă se poate sa îmi dați și exemple multumesc

Answer 1

 

$\displaystyle\bf\\Explicatii~1:\\\\x\Big|\Big(x+3\Big)~~care~inseamna: x~divide~pe~\Big(x+3\Big), este ~echivalent~cu\\\\\Big(x+3\Big)~\vdots~x~~care inseamna~\Big(x+3\Big)~este~divizibil~cu~x.\\\\Se~mai~poate~scrie~asa:~~\frac{x+3}{x}\in N~adica~impartirea~este~fara~rest.\\\\Explicatia~2\\\\Metoda~de~rezolvare: \\\\Scriem~fractia~(ca~mai~sus)~pe~care~o~descompunem\\~in~suma~de~2~fractii~din~care~prima~prin~simplificare\\devine~un~numar~natural~si~a~doua~fractie~este~mai~usor~de~rezolvat.$

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$\displaystyle\bf\\Rezolvare:\\a)\\x\Big|\Big(x+3\Big)~~\Longleftrightarrow~~\frac{x+3}{x}\in N\\\\\frac{x+3}{x}=\frac{x}{x}+\frac{3}{x}=1+\frac{3}{x}\\\\\frac{x}{x}=1\in N\\\\\implies \frac{3}{x}\in N\\\\\implies x\in D_3\\\\D_3=\{1;~3\}\\\\\boxed{\bf x\in\{1;~3\}}$

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$\displaystyle\bf\\b)\\\Big(2x\Big)\Big|\Big(x+5\Big)\\\\\Big(x+5\Big)\vdots\Big(2x\Big)\implies \Big(2(x+5)\Big)\vdots\Big(2x\Big)\\\\\frac{2(x+5)}{2x}\in N\\\\\frac{2x+10}{2x}=\frac{2x}{2x}+\frac{10}{2x}\\\\\frac{2x}{2x}=1\in N~\implies~\frac{10}{2x}\in N\\\\\Big(2x\Big)\in D_{10}\\D_{10}=\{1;~2;~5;~10\}\\\\2x=1~\implies x=\frac{1}{2} \notin N\\\\2x=2~\implies x=\frac{2}{2}=1 \in N\\\\2x=5~\implies x=\frac{5}{2} \notin N\\\\2x=10~\implies x=\frac{10}{2}=5 \in N\\\\\boxed{\bf x\in\{1;~5\}}$

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$\displaystyle\bf\\\\c)\\\Big(x+2\Big)\Big|\Big(2x+1\Big)\\\\\Big(2x+1\Big)\vdots\Big(x+2\Big)\\\\\frac{2x+1}{x+2}\in N\\\\\frac{2x+1}{x+2}=\frac{2x+4-4+1}{x+2}=\frac{2x+4-3}{x+2}=\\\\=\frac{2x+4}{x+2}-\frac{3}{x+2}=\frac{2(x+2)}{x+2}-\frac{3}{x+2}=\left(2-\frac{3}{x+2}\right)\in N\\\\\Big(x+2\Big)\in D_3\\\\D_3=\{1;~3\}\\\\x+2=1~\implies x=1-1=(-1)\notin N\\\\x+2=3~\implies x=3-1=1\in N\\\\\boxed{\bf x=1}$

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$\displaystyle\bf\\d)\\\Big(3x+1\Big)\Big|\Big(2x+5\Big)\\\\\Big(2x+5\Big)\vdots\Big(3x+1\Big)\implies \Big(3(2x+5)\Big)\vdots\Big(3x+1\Big)\\\\\frac{3(2x+5)}{3x+1}=\frac{6x+15}{3x+1}=\frac{6x+2+13}{3x+1}=\frac{6x+2}{3x+1}+\frac{13}{3x+1}\\\\\frac{6x+2}{3x+1}=\frac{2(3x+1)}{3x+1}=2\in N\\\\\implies\frac{13}{3x+1}\in N\\\\\Big(3x+1\Big)\in D_{13}\\\\D_{13}=\{1;~13\}\\\\3x+1=1 \implies x=0 \in N\\\\3x+1=13 \implies x=\frac{13-1}{3}=4 \in N\\\\\boxed{\bf x\in \{0;~4\}}$