Question

determinai toate valorie intregi ale lui x , astfel incat $17/2x +5 sa fie nr intreg$

Answer 1

$\displaystyle\dfrac{17}{2x+5}\in Z\implies 2x+5|17\implies 2x+5\in D_{17} \\ \implies 2x+5\in\{-17 \ ; \ -1 \ ; \ 1 \ ; \ 17\}|-5 \\ 2x\in\{-22 \ ; \ -6 \ ; \ -4 \ ; \ 12\}|:2 \\ x\in\{-11 \ ; \ -3 \ ; \ -2 \ ; \ 6\}$

Answer 2

17/(2x+5)€ Z
=> (2x+5)€ divizorilor lui 17
(2x+5)€ {-17;-1;1;17} |-5
2x€ {-22;-6;-4;12} |:2
x€ {-11;-3;-2,6}