Để B \(\in\)Z
=> \(x+1⋮x+5\)
=> \(x+5-4⋮x+5\)
Ta có : Vì \(x+5⋮x+5\)
=> \(-4⋮x+5\)
=> \(x+5\in-4\)
=> \(x+5\in\left\{\pm1;\pm2;\pm4\right\}\)
Lập bảng xét 6 trường hợp
| \(x+5\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(4\) | \(-4\) |
| \(x\) | \(-4\) | \(-6\) | \(-3\) | \(-7\) | \(-1\) | \(-9\) |
Vậy \(B\inℤ\Leftrightarrow x\in\left\{-4;-6;-3;-7;-1;-9\right\}\)
\(B=\frac{x+1}{x+5}=\frac{x+5-4}{x+5}=1-\frac{4}{x+5}.\)
Để \(B\in Z\Leftrightarrow\frac{4}{x+5}\in Z\)\(\Rightarrow4\)\(⋮\)\(x+5\)
\(\Rightarrow x+5\inƯ_4\)Mà \(Ư_4=\left\{\pm1;\pm2;\pm4\right\}\)
TH1 : \(x+5=1\Rightarrow x=-4\)
Th2 : \(x+5=-1\Rightarrow x=-6\)
TH3 : \(x+5=2\Rightarrow x=-3\)
TH4 : \(x+5=-2\Rightarrow x=-7\)
TH5 : \(x+5=4\Rightarrow x=-1\)
TH6 : \(x+5=-4\Rightarrow x=-9\)
\(KL:x\in\left\{-4;-6;-3;-7;-1;-9\right\}\)
Để B \(\in Z\)
\(x+1⋮x+5\)\(\Leftrightarrow x+5-4⋮x+5\)
Vì
\(x+5⋮x+5\)
\(\Rightarrow-4⋮x+5\)
\(\Rightarrow x\in\left(-4\right)=\left\{\mp1;\mp2;\mp4\right\}\)
Ta có bảng
| x+5 | -1 | 1 | -2 | 2 | -4 | 4 |
| x | -4 | -6 | -7 | -3 | -9 | -1 |
Vậy x = -4;-6;-7;-3;-9;-1
