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