Giúp mình với mọi người ơi

Giải thích các bước giải:

a.ĐKXĐ: $x\ne\pm1$

Ta có:

$A=(\dfrac{x+x^3}{1-x^2}-\dfrac{x-x^3}{1+x^2}):(\dfrac{1+x}{1-x}-\dfrac{1-x}{1+x})$

$\to A=(-\dfrac{x+x^3}{x^2-1}-\dfrac{x-x^3}{x^2+1}):(-\dfrac{1+x}{x-1}-\dfrac{1-x}{1+x})$ 

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

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

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

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

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

b.ĐKXĐ: $x\ne \pm1$

Ta có:

$B=(\dfrac{5x+1}{x^3-1}-\dfrac{1-2x}{x^2+x+1}-\dfrac2{x-1}):\dfrac{2x}{x^2-1}$

$\to B=(\dfrac{5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(1-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)})\cdot \dfrac{x^2-1}{2x}$

$\to B=\dfrac{5x+1-\left(1-2x\right)\left(x-1\right)-2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot \dfrac{x^2-1}{2x}$

$\to B=0\cdot \dfrac{x^2-1}{2x}$

$\to B=0$