Ta có: 3(x + 4) + 4(x + 3) = 3x + 12 + 4x + 12 = 7x + 24 = 7(x + 2) + 10
Do : 7(x + 2) \(⋮\)x + 2
Để 7(x + 2) + 10 \(⋮\)x + 2 thì 10 \(⋮\)x + 2 => x + 2 \(\in\)Ư(10) = {1; 2; 5; 10}
Lập bảng :
| x + 2 | 1 | 2 | 5 | 10 |
| x | -1(loại) | 0 | 3 | 8 |
Vậy ....
\(3\left(x+4\right)+4\left(x+3\right)=3x+12+4x+12\)
\(=7x+24=7x+14+10=7\left(x+2\right)+10\)
\(\Rightarrow3\left(x+4\right)+4\left(x+3\right)\)\(⋮\)\(x+2\)\(\Leftrightarrow x+2\inƯ_{10}\)
Mà \(Ư_{10}=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)\(\Rightarrow...\)
\(3\left(x+4\right)+4\left(x+3\right)⋮x+2\)
\(\Rightarrow3x+12+4x+12⋮x+2\)
\(\Rightarrow7x+24⋮x+2\)
\(\Rightarrow7x+14+10⋮x+2\)
\(\Rightarrow2\left(x+2\right)+10⋮x+2\)
Ta có:\(2\left(x+2\right)⋮x+2\Rightarrow10⋮x+2\)
\(\Rightarrow x+2\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(x\inℕ\Rightarrow x+2\ge2\Rightarrow x+2\in\left\{2;5;10\right\}\)
\(x\in\left\{0;3;7\right\}\)
Vậy\(x=0;3;7\)
Ta có : 3(x + 4) + 4(x + 3) \(⋮\)x + 2
=> 3x + 12 + 4x + 12 \(⋮\)x + 2
=> 3x + 4x + 12 + 12 \(⋮\)x + 2
=> 7x + 24 \(⋮\)x + 2
=> 7x + 14 + 10 \(⋮\)x + 2
=> 7(x + 2) + 10 \(⋮\)x + 2
Ta có : Vì \(7\left(x+2\right)⋮x+2\)
=> \(10⋮x+2\)
=> \(x+2\inƯ\left(10\right)\)
=> \(x+2\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Lập bảng xét các trường hợp ta có :
| \(x+2\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(5\) | \(-5\) | \(10\) | \(-10\) |
| \(x\) | \(-1\) | \(-3\) | \(0\) | \(-4\) | \(3\) | \(-7\) | \(8\) | \(-12\) |
Vậy \(3\left(x+4\right)+4\left(x+3\right)⋮x+2\)\(\Leftrightarrow x\in\left\{-1;-3;0;-4;3;-7;8;-12\right\}\)
