Question

Answer 1

b: Để A nguyên thì \(x+3\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{-4;-1;-5;1;-7\right\}\)

Answer 2

\(a,A=\dfrac{x+x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{x+3}=\dfrac{2x+2}{x+3}\\ b,A=\dfrac{2\left(x+3\right)-4}{x+3}=2-\dfrac{4}{x+3}\in Z\\ \Leftrightarrow x+3\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow x\in\left\{-7;-5;-4;-1;1\right\}\left(x\ne-2\right)\)