Question

Help mee!!!!Va rog !!!

Answer 1

$\left. \begin{aligned} \displaystyle\frac{x+3}{3x-1}\in\mathbb{Z}\Leftrightarrow 3x-1/x+3\Rightarrow 3x-1/3(x+3)&\Rightarrow 3x-1/3x+9\\ &\text{Cum } 3x-1/3x-1 \end{aligned}\right\}\\ \Rightarrow 3x-1/(3x+9)-(3x-1)\Rightarrow 3x-1/10\Rightarrow 3x-1\in D_{10}\\D_{10}=\{\pm1,\pm2,\pm5,\pm10\}\\3x-1\in\{-10, -5, -2, -1, 1, 2, 5, 10\}|+1\\3x\in\{-9, -4, -1, 0, 2, 3, 6, 11\}|:3\\x\in\left\{-3, -\displaystyle\frac43, -\frac13, 0, \frac23, 1, 2, \frac{11}3\right\}$

$x\in\mathbb{Z}\Rightarrow x \in\{-3, 0, 1, 2\}$