a) Ta có: \(n+3⋮n-1\)
\(\Leftrightarrow n-1+4⋮n-1\)
mà \(n-1⋮n-1\)
nên \(4⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)
Vậy: \(n\in\left\{2;0;3;-1;5;-3\right\}\)
b) Ta có: \(4n+3⋮2n+1\)
\(\Leftrightarrow4n+2+1⋮2n+1\)
mà \(4n+2⋮2n+1\)
nên \(1⋮2n+1\)
\(\Leftrightarrow2n+1\inƯ\left(1\right)\)
\(\Leftrightarrow2n+1\in\left\{1;-1\right\}\)
\(\Leftrightarrow2n\in\left\{0;-2\right\}\)
hay \(n\in\left\{0;-1\right\}\)
Vậy: \(n\in\left\{0;-1\right\}\)
Đúng 0
Bình luận (0)
