a)
\(n+5⋮n+1\)
\(\Rightarrow n+1+4⋮n+1\)
\(\Rightarrow4⋮n+1\Rightarrow n+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{0;-2;1;-3;3;-5\right\}\)
\(a,\left(n+5\right)⋮\left(n+1\right)\Leftrightarrow\left(n+1\right)+4⋮\left(n+1\right)\)
\(\Leftrightarrow4⋮n+1\left(n\inℤ\right)\)
\(\Leftrightarrow n+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow n=-2;0;-3;1;-5;3\)
Vậy \(n=-5;-3;-2;0;1;3\)
a) Ta có: \(\frac{n+5}{n+1}=\frac{n+1+4}{n+1}=1+\frac{4}{n+1}\)
Để (n+5) chia hết cho (n+1)
Thì 4 phải chia hết cho n+1
\(\Rightarrow n+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Lập bảng ta có:
| n+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| n | 0 | -2 | 1 | -3 | 3 | -5 |
Vậy số nguyên n thỏa mãn là
n = {-5;-3;-2;0;1;3}
b) \(\frac{6n+4}{2n+1}=\frac{3\left(2n+1\right)+1}{2n+1}=3+\frac{1}{2n+1}\)
Để (6n+4) chia hết cho (2n+1)
Thì 1 phải chia hết cho 2n+1
\(\Rightarrow2n+1\inƯ\left(1\right)=\left\{1;-1\right\}\)
| 2n+1 | 1 | -1 |
| n | 0 | -1 |
Vậy n = {-1;0}
\(\left(6n+4\right)⋮\left(2n+1\right)\)
\(\Rightarrow3.\left(2n+1\right)+1⋮\left(2n+1\right)\)
\(\Rightarrow1⋮\left(2n+1\right)\Rightarrow\left(2n+1\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow n\in\left\{0;-1\right\}\)
\(b,\left(6n+4\right)⋮\left(2n+1\right)\Leftrightarrow\left(6n+3+1\right)⋮\left(2n+1\right)\)
\(\Leftrightarrow3\left(2n+1\right)+1⋮\left(2n+1\right)\)
\(\Leftrightarrow1⋮2n+1\left(2n+1\inℤ\right)\)
\(\Leftrightarrow2n+1\inƯ\left(1\right)=\left\{-1;1\right\}\)
\(\Leftrightarrow n\in\left\{-1;0\right\}\)
Vậy \(n=-1;0\)
