ĐKXĐ : \(x\ne0;\pm2\)
Ta có :
\(A=\left(\frac{x+2}{x-2}-\frac{4x}{4-x^2}-\frac{x-2}{x+2}\right):\frac{x^3+x^2+2x}{x-2}\)
\(=\left(\frac{x+2}{x-2}+\frac{4x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{x+2}\right):\frac{x\left(x^2+x+2\right)}{x-2}\)
\(=\left(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{4x}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right)\) \(.\frac{x-2}{x\left(x^2+x+2\right)}\)
\(=\frac{x^2+2x+4+4x-x^2+2x-4}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{x\left(x^2+x+2\right)}\)
\(=\frac{x^2+8x}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{x\left(x^2+x+2\right)}\)
\(=\frac{x+8}{\left(x+2\right)\left(x^2+x+2\right)}\)
Vậy...
b/ Em tìm x, đối chiếu điều kiện, rồi thay vào A vừa rút gọn đc là xong nhé !