Xét phân số \(A=\dfrac{x+2}{x+3}\). x + 2 chia hết cho x + 3 khi A \(\in Z\)
\(A=\dfrac{x+2}{x+3}=\dfrac{x+3-1}{x+3}=\dfrac{x+3}{x+3}-\dfrac{1}{x+3}=1-\dfrac{1}{x+3}\)
\(A\in Z\) => x + 3 \(\inƯ\left(1\right)=\left\{-1,1\right\}\)
x + 3 = -1 <=> x = -4
x + 3 = 1 <=> x = -2
Vậy ....