Các anh chị giúp em với ạ

Giải thích các bước giải:

a.Ta có:

$A=(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3})\cdot (1-\dfrac1x-\dfrac2{x^2})$

$\to A=(\dfrac{x^2-2x}{2(x^2+4)}-\dfrac{2x^2}{-(x-2)(x^2+4)})\cdot \dfrac{x^2-x-2}{x^2}$

$\to A=(\dfrac{x^2-2x}{2(x^2+4)}+\dfrac{2x^2}{(x-2)(x^2+4)})\cdot \dfrac{(x-2)(x+1)}{x^2}$

$\to A=(\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x-2\right)\left(x^2+4\right)}+\dfrac{4x^2}{2\left(x-2\right)\left(x^2+4\right)})\cdot \dfrac{(x-2)(x+1)}{x^2}$

$\to A=(\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)})\cdot \dfrac{(x-2)(x+1)}{x^2}$

$\to A=(\dfrac{x^3+4x}{2\left(x-2\right)\left(x^2+4\right)})\cdot \dfrac{(x-2)(x+1)}{x^2}$

$\to A=\dfrac{x}{2\left(x-2\right)}\cdot \dfrac{(x-2)(x+1)}{x^2}$

$\to A=\dfrac{x+1}{2x}$

b.Để $A=\dfrac52$

$\to \dfrac{x+1}{2x}=\dfrac52$

$\to2x+2=10x$

$\to 8x=2$

$\to x=\dfrac14$