Giải:
Có:
\(2x+4=2x+2+2=2\left(x+1\right)+2\)
Vì \(2\left(x+1\right)+2⋮x+1\)
Mà \(2\left(x+1\right)⋮x+1\)
\(\Leftrightarrow2⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(2\right)\)
\(\Leftrightarrow x+1\in\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{-3;-2;0;1\right\}\) (Thỏa mãn)
Vậy \(x\in\left\{-3;-2;0;1\right\}\).
Chúc bạn học tốt!!!
\(2x+4⋮x+1\)
\(=\)\(2\left(x+1\right)+2⋮x+1\)
\(\Rightarrow2⋮x+1\)
Bảng:
| x + 1 | -1 | 1 | -2 | 2 |
| x | -2 | 0 | -3 | 1 |
Ta có:
\(\dfrac{2x+4}{x+1}=\dfrac{2x+4}{2x+2}=\dfrac{2x+2+2}{2x+2}=1+\dfrac{2}{2x+2}\)
\(=1+\dfrac{1}{x+1}\)
Để \(\dfrac{2x+4}{x+1}\) đạt giá trị nguyên thì \(\dfrac{1}{x+1}\) đạt giá trị nguyên.
\(\Rightarrow x+1\inƯ\left(1\right)\)
\(\Rightarrow x+1\in\left\{-1;1\right\}\Rightarrow x\in\left\{-2;0\right\}\)
Vậy................
Chúc bạn học tốt!!!
Ta có: \(2x+4=2x+2+2=2\left(x+1\right)+2\)
Để \(2x+4⋮x+1\) thì \(2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng sau:
| x+1 | -2 | -1 | 1 | 2 |
| x | -3 | -2 | 0 | 1 |
Vì \(x\in Z\) nên \(x\in\left\{-3;-2;0;1\right\}\)
Bài làm
Ta có : 2x+4 \(⋮\) x + 1 \(\in\)
= 2x+2+2 \(⋮\) x + 1
= 2 .(x+1)+2 \(⋮\) x + 1
Mà 2 ( x + 1 ) \(⋮\) x+1
=> 2 \(⋮\) x + 1
=> x+1\(\in\) Ư (2)
=> x + 1 \(\in\)\(\left\{\pm1;\pm2\right\}\)
Vậy x \(\in\left\{-2;-3;0;1\right\}\)
