vì x + 16 ⋮ x + 1
x+1⋮ x+1
=>(x+16)-(x+1)\(⋮x+1\)
=>\(15⋮x+1\)
=>(x+1)\(\inƯ\left(15\right)\) ={\(\pm1;\pm3;\pm5;\pm15\) }
ta có bảng
| x+1 | -1 | 1 | -3 | 3 | -5 | 5 | -15 | 15 |
| x | -2 | 0 | -4 | 2 | -6 | 4 | -16 | 14 |
vậy x\(\in\left\{-16;-6;-4;-2;0;2;4;14\right\}\)
x + 11 x + 1
vì \(x+1⋮x+1\)
=>\(\left(x+11\right)-\left(x+1\right)⋮\left(x+1\right)\)
=>\(\left(x+11-0x-1\right)⋮\left(x+1\right)\)
=> \(10⋮x+1\)
=>\(x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
ta có bảng sau:
| x+1 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
| x | -11 | -6 | -3 | -2 | 0 | 1 | 4 |
9 |
vậy\(x\in\left\{-11;-6;-3;-2;0;1;4;9\right\}\)
a) (x + 16) (x + 1)
Ta có: x + 16 = (x + 1) +15 nên (x + 1) + 15 ⋮ (x + 1) khi 15 \(⋮\) (x + 1)
\(\Rightarrow\) x + 1 \(\in\) Ư(15) = {1; 3; 5;15}
\(\Rightarrow\) x x \(\in\) {0; 2; 4; 14}
Vậy x \(\in\) {0; 2; 4; 14}.
b) (x + 11) x + 1)
Ta có: x + 11 = (x + 1) + 10 nên (x + 1) + 10 ⋮ (x + 1) khi 10 \(⋮\) (x + 1)
\(\Rightarrow\) x + 1 \(\in\) Ư(10) = {1; 2; 5; 10}
\(\Rightarrow\) x \(\in\) {0; 1; 4; 9}
Vậy x \(\in\) {0; 1; 4; 9}.
