Ta lm 1 bài còn bài thứ 2 tương tự
a) A = \(\dfrac{2x+4}{x+3}\) = \(\dfrac{2\left(x+3\right)-2}{x+3}\) = \(2-\dfrac{2}{x+3}\)
Để A \(\in\) Z => \(2-\dfrac{2}{x+3}\in Z\)
Mà 2 \(\in\) Z => \(\dfrac{2}{x+3}\) \(\in\) Z
=> 2 \(⋮\) \(x+3\)
=> \(x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> x + 3 = 1 => x = -2 => A = -4 (TM)
x + 3 = -1 x = -4 => A = 4 ( TM)
x + 3 = 2 x = -1 => A = 1 (TM)
x+ 3 = -2 x = -5 => A = 3 (TM)