Question

$Aflati~~~ A ={{x\in Z | \sqrt[3]{ \frac{7x+2}{x+5}}\in Z } }$

Answer 1

$\dfrac{7x+2}{x+5} = \dfrac{7x+35-33}{x+5} = \dfrac{7(x+5)}{x+5}-\dfrac{33}{x+5} = 7-\dfrac{33}{x+5} \\ \\ x+5 | 33 \Rightarrow x+5 \in D_{33} \Rightarrow x+5 \in \Big\{-33,-11,-1,1,11,33\Big\}\Big|^{-1} \Rightarrow \\ \\ \Rightarrow \dfrac{1}{x+5}\in \Big\{-\dfrac{1}{33},-\dfrac{1}{11},-1,1,\dfrac{1}{11},\dfrac{1}{33}\Big\}\Big|\cdot (-33) \Rightarrow \\ \\ \Rightarrow -\dfrac{33}{x+5}\in \Big\{1,3,33,-33,-3,-1\Big\}\Big|+7 \Rightarrow$

$\Rightarrow 7-\dfrac{33}{x+5}\in \Big\{8,10,40,26,4,6\Big\} \\ \\ Extragem cuburile perfecte: \\ \\ \Rightarrow 7-\dfrac{33}{x+5} = 8 \Rightarrow \dfrac{33}{x+5} = 7-8 \Rightarrow \dfrac{33}{x+5} = -1 \Rightarrow 33=-x-5 \Rightarrow \\ \\ \Rightarrow x = -5-33 \Rightarrow \boxed{x = -38}\rightarrow solutie~unica.$

$\Rightarrow \boxed{A = \Big\{-38\Big\}}$