Răspuns:
Explicație pas cu pas:
[tex]c) A=\{x\in\mathbb{N}|\frac{7x+9}{2x+1}\in\mathbb{N}\}\\2x+1|7x+9\Rightarrow 2x+1|14x+18\\2x+1|2x+1\Rightarrow 2x+1|14x+7\\\texttt{De aici rezulta ca:} 2x+1|(14x+18-14x-7)\Rightarrow 2x+1|11\\2x+1\in D_{11}\\2x+1\in \{1,11\}\\2x\in \{0,10\}\\x\in \{0,5\}\\A=\{0,5\}[/tex]
[tex]d)A=\{x\in\mathbb{Z}|\frac{5x+13}{2x+1}\in\mathbb{Z}\}\\2x+1|5x+13\Rightarrow 2x+1|10x+26\\2x+1|2x+1\Rightarrow 2x+1|10x+5\\\texttt{De aici rezulta ca }2x+1 | (10x+26-10x-5)\Rightarrow 2x+1|21\\2x+1\in D_{21}\\2x+1\in \{-21,-7,-3,-1,1,3,7,21\}\\2x\in \{-22,-8,-4,-2,0,2,6,20\}\\x\in\{-11,-4,-2,-1,0,1,3,10\}\\A={-11,-4,-2,-1,0,1,3,10\}[/tex]
[tex]\texttt{Am folosit formulele: }\\\bullet a|b\Rightarrow a|b\cdot k\\\bullet a|b \texttt{ si } a|c\Rightarrow a|(b-c)[/tex]
