a) Ta có:
\(n+5⋮n+1\)
\(\Rightarrow\left(n+1\right)+4⋮n+1\)
\(\Rightarrow4⋮n+1\)
\(\Rightarrow n+1\in U\left(4\right)=\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow n\in\left\{-2;0;-3;1;-5;3\right\}\)
Vậy \(n\in\left\{-2;0;-3;1;-5;3\right\}\)
b) Ta có:
\(4n+3⋮2n-1\)
\(\Rightarrow\left(4n-2\right)+5⋮2n-1\)
\(\Rightarrow2\left(2n-1\right)+5⋮2n-1\)
\(\Rightarrow5⋮2n-1\)
\(\Rightarrow2n-1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow n\in\left\{0;1;-2;3\right\}\)
Vậy \(n\in\left\{0;1;-2;3\right\}\)
a) (n + 5) chia hết cho n + 1.
Ta có: n + 5 = (n + 1) + 4 \(\Rightarrow\) (n + 1) + 4 \(⋮\) (n + 1) khi 4 \(⋮\) (n + 1)
\(\Rightarrow\) n + 1 \(\in\) Ư(4) = {1; 2; 4}
\(\Rightarrow\) n \(\in\) {0; 1; 3}
Vậy n \(\in\) {0; 1; 3}.
b) (4n + 3) chia hết cho 2n - 1.
Vì (2n - 1) \(⋮\) (2n - 1)
\(\Rightarrow\) (4n - 2) \(⋮\) (2n - 1)
Mà (4n + 3) \(⋮\) (2n + 1)
\(\Rightarrow\) (4n + 5 - 2) \(⋮\) (2n + 1)
\(\Rightarrow\) (4n + 5 - 2) - (4n - 2) \(⋮\) (2n - 1)
\(\Rightarrow\) 5 \(⋮\) (2n - 1)
\(\Rightarrow\) 2n - 1 \(\in\) Ư(5) = {1; 5}
\(\Rightarrow\) 2n \(\in\) {0; 4}
\(\Rightarrow\) n \(\in\) {0; 2}
Thử lại: n = 0 \(\Rightarrow\) 3 \(⋮\) 1
n = 2 \(\Rightarrow\) 11 \(⋮̸\) 4
Vậy n = 0 (TM).
