1) \(A=\dfrac{4}{2x+1}\in Z\)\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)Do \(x\in Z\)\(\Rightarrow x\in\left\{-1;0\right\}\)2) \(B=\dfrac{x+1}{x-2}=1+\dfrac{3}{x-2}\in Z\)\(\Rightarrow x-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)\(\Rightarrow x\in\left\{-1;1;3;5\right\}\)3) \(C=\dfrac{2x-1}{x-3}=2+\dfrac{5}{x-3}\in Z\)\(\Rightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)\(\Rightarrow x\in\left\{-2;2;4;8\right\}\)4) \(D=\dfrac{6x+7}{2x+3}=\dfrac{3\left(2x+3\right)}{2x+3}-\dfrac{2}{2x+3}=3-\dfrac{2}{2x+3}\in Z\)\(\Rightarrow2x+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)Do \(x\in Z\)\(\Rightarrow x\in\left\{-2;-1\right\}\) Câu trả lời: 1) \(A=\dfrac{4}{2x+1}\in Z\)\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)Do \(x\in Z\)\(\Rightarrow x\in\left\{-1;0\right\}\)2) \(B=\dfrac{x+1}{x-2}=1+\dfrac{3}{x-2}\in Z\)\(\Rightarrow x-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)\(\Rightarrow x\in\left\{-1;1;3;5\right\}\)3) \(C=\dfrac{2x-1}{x-3}=2+\dfrac{5}{x-3}\in Z\)\(\Rightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)\(\Rightarrow x\in\left\{-2;2;4;8\right\}\)4) \(D=\dfrac{6x+7}{2x+3}=\dfrac{3\left(2x+3\right)}{2x+3}-\dfrac{2}{2x+3}=3-\dfrac{2}{2x+3}\in Z\)\(\Rightarrow2x+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)Do \(x\in Z\)\(\Rightarrow x\in\left\{-2;-1\right\}\)