Giải thích các bước giải:
$A=\dfrac{x^4-x}{x^2+x+1}-\dfrac{2x^2+x}{x}+\dfrac{2(x^2-1)}{x-1}$ $\text{(ĐK:$x\neq0;x\neq1$)}$
$=\dfrac{x(x^3-1)}{x^2+x+1}-\dfrac{x(2x+1)}{x}+\dfrac{2(x-1)(x+1)}{x-1}$
$=\dfrac{x(x-1)(x^2+x+1)}{x^2+x+1}-(2x+1)+2(x+1)$
$=x(x-1)-2x-1+2x+2$
$=x^2-x+1$
Học tốt!!!