Đáp án:
$n=\{\dfrac{19}{6};\dfrac{-2}{3};\dfrac{4}{3};\dfrac{7}{6}\}$
Giải thích các bước giải:
$\dfrac{3n+2}{4n-5}$
$→ 3n+2 \quad\vdots\quad 4n-5$
$→ 12n+8 \quad\vdots\quad 12n-15$
$→ 12n - 15 + 23 \quad\vdots\quad 12n-15$
$→ 23 \quad\vdots\quad 12n-15$
$→ 12n-15 \in Ư(23)=\{±1;±23\}$
Ta có bảng sau:
\begin{array}{|c|c|}\hline 12n-15&23&-23&1&-1\\\hline n&\dfrac{19}{6}&\dfrac{-2}{3}&\dfrac{4}{3}&\dfrac{7}{6}\\\hline\end{array}
Vậy $n=\{\dfrac{19}{6};\dfrac{-2}{3};\dfrac{4}{3};\dfrac{7}{6}\}$
