c, Ta có :
\(5n+3⋮4n+1\)
Mà \(4n+1⋮4n+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}20n+12⋮4n+1\\20n+5⋮4n+1\end{matrix}\right.\)
\(\Leftrightarrow7⋮4n+1\)
Vì \(n\in Z\Leftrightarrow4n+1\in Z;4n+1\inƯ\left(7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4n+1=7\\4n+1=1\\4n+1=-7\\4n+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=\dfrac{3}{2}\left(loại\right)\=0\=-2\=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Vậy -
d, Ta có :
\(6n-7⋮3n+2\)
Mà \(3n+2⋮3n+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}6n-7⋮3n+2\\6n+4⋮3n+2\end{matrix}\right.\)
\(\Leftrightarrow11⋮3n+2\)
Vì \(n\in Z\Leftrightarrow3n+2\in Z;3n+2\inƯ\left(11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3n+2=11\\3n+2=1\\3n+2=-11\\3n+2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=3\=-\dfrac{1}{3}\=\dfrac{-13}{3}\=-1\end{matrix}\right.\)
Vậy -
