Đáp án: $ x\in\{2,3,4,1,5,0,6,-1,9,-4\}$
Giải thích các bước giải:
Để $B\in Z$
$\to \dfrac{2x+3}{x^2-5x-1}\in Z$
$\to 2x+3\quad\vdots\quad x^2-5x-1$
$\to (2x+3)(2x-13)\quad\vdots\quad x^2-5x-1$
$\to 4x^2-20x-39\quad\vdots\quad x^2-5x-1$
$\to (4x^2-20x-4)-35\quad\vdots\quad x^2-5x-1$
$\to 4(x^2-5x-1)-35\quad\vdots\quad x^2-5x-1$
$\to 35\quad\vdots\quad x^2-5x-1$
$\to x^2-5x-1\in U(35)$
Mà $x^2-5x-1=(x-\dfrac52)^2-\dfrac{29}{4}\ge -\dfrac{29}{4}$
$\to x^2-5x-1\in\{-7,-5,-1,1,5,7,35\}$
Vì $x\in Z$
$\to x\in\{2,3,4,1,5,0,6,-1,9,-4\}$
