Giải thích các bước giải:
$\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+x^2+1)}$
$\to \dfrac{(x+1)(x^2-x+1)-(x-1)(x^2+x+1)}{(x^2-x+1)(x^2+x+1)}=\dfrac{3}{x(x^4+x^2+1)}$
$\to \dfrac{x^3+1-(x^3-1)}{(x^2+1)^2-x^2}=\dfrac{3}{x(x^4+x^2+1)}$
$\to \dfrac{2}{x^4+x^2+1}=\dfrac{3}{x(x^4+x^2+1)}$
$\to 2=\dfrac{3}{x}$
$\to x=\dfrac 32$