Bạn sửa lại đề dùm mình nha, sai đề hơi nhiều đó.
ĐKXĐ:\(x\ne0;2\)
\(P=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2+2x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\\ P=\left(\frac{x\left(x-2\right)}{2\left(x^2+4\right)}-\frac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right).\frac{x^2-x-2}{x^2}\\ P=\left(\frac{x\left(x-2\right)}{2\left(x^2+4\right)}+\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right).\frac{x^2-2x+x-2}{x^2}\\ P=\left(\frac{x\left(x-2\right)^2}{2\left(x^2+4\right)\left(x-2\right)}+\frac{4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right).\frac{x\left(x-2\right)+\left(x-2\right)}{x^2}\)
\(P=\frac{x\left(x^2-4x+4\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\frac{x^3-4x^2+4x-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{x^2}\\ P=\frac{\left(x^3+4x\right)\left(x-2\right)\left(x+1\right)}{2\left(x^2+4\right)\left(x-2\right).x^2}\\ P=\frac{x\left(x^2+4\right)\left(x-2\right)\left(x+1\right)}{2x^2\left(x^2+4\right)\left(x-2\right)}\\ P=\frac{x+1}{2x}\)
