b, \(A=\dfrac{x+3+2}{x+3}=1+\dfrac{2}{x+3}\Rightarrow x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+3 | 1 | -1 | 2 | -2 |
x | -2 | -4 | -1 | -5 |
a, Để A là phân số thì \(x+3\ne0\Leftrightarrow x\ne-3\)
b, \(A=\dfrac{x+5}{x+3}=\dfrac{x+3+2}{x+3}=1+\dfrac{2}{x+3}\)
\(\Rightarrow x+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Ta có bảng:
x+3 | -2 | -1 | 1 | 2 |
x | -5 | -4 | -2 | -1 |
Vậy \(x\in\left\{-5;-4;-2;-1\right\}\)