Question

Rút gọn biểu thức sau

\(\left(\dfrac{x}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)

Answer 1

điều kiện xác định : \(x\ne\pm2\)

ta có : \(\left(\dfrac{x}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)

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

\(=\left(\dfrac{x^2+2x-4}{\left(x+2\right)^2}\right):\left(\dfrac{-x}{\left(x-2\right)\left(x+2\right)}\right)=\left(\dfrac{x^2+2x-4}{\left(x+2\right)^2}\right).\left(\dfrac{-\left(x-2\right)\left(x+2\right)}{x}\right)\)

\(=\dfrac{-\left(x^2+2x-4\right)\left(x-2\right)}{x\left(x+2\right)}\)

đề sai -_-