\(x+11⋮x+1=>x+11-\left(x+1\right)⋮x+1=>10⋮x+1\)
\(=>x+1\inƯ\left(10\right)=\left\{1,2,5,10\right\}\)(vì \(x\in N\))
\(=>x\in\left\{0,1,4,9\right\}\)
Ta có : \(x+11⋮x+1\)
\(\Rightarrow x+1+10⋮x+1\)
Vì \(x+1⋮x+1\)
\(\Rightarrow10⋮x+1\)
\(\Rightarrow x+1\inƯ\left(10\right)\)
\(\Rightarrow x+1\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Lập bảng xét 8 trường hợp
| x + 1 | 1 | - 1 | 2 | - 2 | 5 | - 5 | 10 | - 10 |
| x | 0 (tm) | - 2 (loại) | 1(tm) | - 3 (loại) | 4 (tm) | - 6 (loại) | 9 (tm) | - 11 (loại) |
Vậy \(x\in\left\{0;1;4;9\right\}\)thì \(x+11⋮x+1\)
\(x+11⋮x+1\)
<=> \(x+1+10⋮x+1\)
Mà \(x+1⋮x+1\)
<=> \(10⋮x+1\)
<=> \(x+1\inƯ\left(10\right)=\left\{\pm10;\pm5;\pm2;\pm1\right\}\)
Ta có:
x + 1 = -10 => x = -11
x + 1 = -5 => x = -6
x + 1 = -2 => x = -3
x + 1 = -1 => x = -2
x + 1 = 1 => x = 0
x + 1 = 2 => x = 1
x + 1 = 5 => x = 4
x + 1 = 5 => x = 9
Mà \(x\inℕ\)nên \(x\in\left\{0;1;4;9\right\}\)
