Question

tìm n thuộc Z biết 5*n^2+11 chia hết cho 2*n^2+1

Answer 1

\(5n^2+11⋮2n^2+1\)

Mà \(2n^2+1⋮2n^2+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}10n^2+22⋮2n^2+1\\10n^2+5⋮2n^2+1\end{matrix}\right.\)

\(\Leftrightarrow17⋮2n^2+1\)

\(\Leftrightarrow2n^2+1\inƯ\left(17\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2n^2+1=1\\2n^2+1=17\\2n^2+1=-1\\2n^2+1=-17\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=0\left(tm\right)\\\left[{}\begin{matrix}n=\sqrt{8}\left(loại\right)\\n=-\sqrt{8}\left(loại\right)\end{matrix}\right.\\n^2=-1\left(loại\right)\\n^2=-9\left(loại\right)\end{matrix}\right.\)

Vậy ...