a: ĐKXĐ: x<>4; x<>-4; x<>-1b: \(A=\dfrac{3x-12-x+20}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x-4}{x+1}=\dfrac{2x+8}{\left(x+4\right)}\cdot\dfrac{1}{x+1}=\dfrac{2}{x+1}\)c: Để A=1 thì x+1=2=>x=1(nhận)Câu trả lời: a: ĐKXĐ: x<>4; x<>-4; x<>-1b: \(A=\dfrac{3x-12-x+20}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x-4}{x+1}=\dfrac{2x+8}{\left(x+4\right)}\cdot\dfrac{1}{x+1}=\dfrac{2}{x+1}\)c: Để A=1 thì x+1=2=>x=1(nhận)