Question

Tính: 

Answer 1

\(\dfrac{x-2}{x^2+2x}+\dfrac{8}{x^2-4}\left(ĐK:x\ne0;x\ne\pm2\right)\)

\(=\dfrac{x-2}{x\left(x+2\right)}+\dfrac{8}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{\left(x-2\right)^2}{x\left(x+2\right)\left(x-2\right)}+\dfrac{8x}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2-4x+4+8x}{x\left(x+2\right)\left(x-2\right)}\)

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

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

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

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