Câu hỏi của Daisy

a: \(=\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x+2\right)}\)\(=\dfrac{x^2+4x-x+2}{x\left(x-2\right)\left(x+2\right)}\)\(=\dfrac{\left(x+2\right)\left(x+1\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+1}{x\left(x-2\right)}\)Câu trả lời: a: \(=\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x+2\right)}\)\(=\dfrac{x^2+4x-x+2}{x\left(x-2\right)\left(x+2\right)}\)\(=\dfrac{\left(x+2\right)\left(x+1\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+1}{x\left(x-2\right)}\)