Question

Rút gọn : \(\frac{2+x}{2-x}\div\frac{4x^2}{4-4x+x^2}.\left(\frac{2}{2-x}-\frac{8}{8+x^3}.\frac{4-2x+x^2}{2-x}\right)\)

Answer 1

\(\frac{2+x}{2-x}\div\frac{4x^2}{4-4x+x^2}\times\left(\frac{2}{2-x}-\frac{8}{8+x^3}\times\frac{4-2x+x^2}{2-x}\right)\)

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

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

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

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

\(=\frac{x-2}{2x^2}\)