a) Đặt \(A=\frac{3n-13}{n+3}=\frac{3\left(n+3\right)-22}{n+3}=3-\frac{22}{n+3}\)
=> 22 \(⋮\)n + 3 => n + 3 \(\in\)Ư(22) = { \(\pm1;\pm2;\pm11;\pm22\)}
n + 3 | 1 | -1 | 2 | -2 | 11 | -11 | 22 | -22 |
n | -2 | -4 | -1 | -5 | 8 | -14 | 19 | -25 |
b) Đặt \(B=\frac{2n+3}{n-1}=\frac{2\left(n-1\right)+5}{n-1}=2+\frac{5}{n-1}\)
=> 5 \(⋮\)n - 1 => n - 1 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
n - 1 | 1 | -1 | 5 | -5 |
n | 2 | 0 | 6 | -4 |
\(\left(a\right)3n-13⋮n+3\)
\(3n-13=3\left(n+3\right)-22\)
\(=>n+3=Ư\left(22\right)\)
\(n+3=\left\{-22;-11;-2;-1;1;2;11;22\right\}\)
\(=>n=\left\{-25;-14;-5;-4;-2;-1;8;19\right\}\)
\(\left(b\right)2n+3⋮n-1\)
\(2n+3=2\left(n-1\right)+5\)
\(=>n-1=Ư\left(5\right)\)
\(n-1=\left\{-5;-1;1;5\right\}\)
\(=>n=\left\{-4;0;2;6\right\}\)
a) Để \(3n-13⋮n+3\)
=> \(3n+9-22⋮n+3\)
=> \(3\left(n+3\right)-22⋮n+3\)
Vì \(3\left(n+3\right)⋮n+3\)
=> \(-22⋮n+3\)
=> \(n+3\inƯ\left(-22\right)\)
=> \(n+3\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
=> \(n\in\left\{-2;-4;-1;-5;8;-14;19;-25\right\}\)
b) \(2n+3⋮n-1\)
=> \(2n-2+5⋮n-1\)
=> \(2\left(n-1\right)+5⋮n-1\)
=> \(2\left(n-1\right)⋮n-1\)
=> \(5⋮n-1\)
=> \(n-1\inƯ\left(5\right)\)
=> \(n-1\in\left\{1;-1;5;-5\right\}\)
=> \(n\in\left\{2;0;6;-4\right\}\)