\(A=\frac{2x-3}{x+2}=\frac{2x+4-7}{x+2}=\frac{2\left(x+2\right)-7}{x+2}=2-\frac{7}{x+2}\)
Để \(A\in Z\Leftrightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng:
x+2 | 1 | -1 | 7 | -7 |
x | -1 | -3 | 5 | -9 |
Vậy...
\(A=\frac{2x-3}{x+2}=\frac{2x+4-7}{x+2}=\frac{2\left(x+2\right)-7}{x+2}=2+\frac{-7}{\left(x+2\right)}\)
Mà 2 \(\in Z\Rightarrow\frac{-7}{\left(x+2\right)}\in Z\)
\(\Rightarrow x+2\inƯ\left(-7\right)\)
\(\Rightarrow x+2\in\left\{1;-1;7;-7\right\}\)
Ta có bảng giá trị:
x+2 | 1 | -1 | 7 | -7 |
x | -1 | -3 | 5 | -9 |
C | C | C | C |
Vậy x \(\in:\left\{-1;-3;5;-9\right\}\)