Question

Tìm n∈N để:

A= \(\dfrac{8n+193}{4n+3}\)có giá trị là số tự nhiên.

Answer 1

Để phân số \(A=\dfrac{8n+193}{4n+3}\in N\) thì :

\(8n+193⋮4n+3\)

Mà \(4n+3⋮4n+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}8n+193⋮4n+3\\8n+6⋮4n+3\end{matrix}\right.\)

\(\Leftrightarrow187⋮4n+3\)

\(\Leftrightarrow4n+3\inƯ\left(187\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4n+3=1\\4n+3=11\\4n+3=17\\4n+3=187\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-0,5\left(loại\right)\\n=2\left(tm\right)\\n=3,5\left(loại\right)\\n=46\left(tm\right)\end{matrix}\right.\)

Vậy ..

Answer 2

Để \(A=\dfrac{8n+193}{4n+3}\in N.\)

\(\Rightarrow8n+193⋮4n+3.\)

\(\Rightarrow\left(8n+6\right)+187⋮4n+3.\)

\(\Rightarrow2\left(4n+3\right)+187⋮4n+3.\)

mà \(2\left(4n+3\right)⋮4n+3\Rightarrow187⋮4n+3\Rightarrow4n+3\in U_{\left(187\right)}=\left\{1;11;17;187\right\}.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4n+3=1.\\4n+3=11.\\4n+3=17.\\4n+3=187.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=-\dfrac{1}{2}\left(loại\right).\\n=2\left(tm\right).\\n=\dfrac{7}{2}\left(loại\right).\\n=46\left(tm\right).\end{matrix}\right.\)

Vậy \(n\in\left\{2;46\right\}.\)