Để \(\dfrac{8n+193}{4n+3}\) là số tự nhiên thì:
8n+193 chia hết cho 4n+3
=>8n+6+187 chia hết cho 4n+3
=>2(4n+3)+187 chia hết cho 4n+3
=>187 chia hết cho 4n+3
=>4n+3 ∈Ư(187)={1;11;17;187}
Ta có bảng sau:
| 4n+3 | 1 | 11 | 17 | 187 |
| n | loại | loại | loại | 46 |
Vậy x=46
\(\dfrac{8n+193}{4n+3}=\dfrac{8n+6+187}{4n+3}=\dfrac{2\left(4n+3\right)+187}{4n+3}=\dfrac{2\left(4n+3\right)}{4n+3}+\dfrac{187}{4n+3}=2+\dfrac{187}{4n+3}\)\(\dfrac{8n+193}{4n+3}\in N\Rightarrow187⋮4n+3\)
\(\Rightarrow4n+3\inƯ\left(187\right)\)
\(Ư\left(187\right)=\left\{1;11;17;187\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}4n+3=1\Rightarrow4n=-2\Rightarrow n=\dfrac{-1}{2}\left(KTM\right)\\4n+3=11\Rightarrow4n=8\Rightarrow n=2\left(TM\right)\\4n+3=17\Rightarrow4n=14\Rightarrow n=\dfrac{7}{2}\left(KTM\right)\\4n+3=187\Rightarrow4n=184\Rightarrow n=46\left(TM\right)\end{matrix}\right.\)
