Giải thích các bước giải:
a.$E=1+(\dfrac{2x^3+x^2-x}{x^3-1}-\dfrac{2x-1}{x-1})\dfrac{x^2-x}{2x-1}$
$=1+(\dfrac{x(x+1)(2x-1)}{(x-1)(x^2+x+1)}-\dfrac{2x-1}{x-1})\dfrac{x(x-1)}{2x-1}$
$=1+\dfrac{x(x+1)(2x-1)}{(x-1)(x^2+x+1)}.\dfrac{x(x-1)}{2x-1}-\dfrac{2x-1}{x-1}.\dfrac{x(x-1)}{2x-1}$
$=1+x.\dfrac{x(x+1)}{x^2+x+1}-x$
$=1+x(\dfrac{x^2+x}{x^2+x+1}-1)$
$=1+x\dfrac{x^2+x-x^2-x-1}{x^2+x+1}$
$=1+x\dfrac{-1}{x^2+x+1}$
$=1-\dfrac{x}{x^2+x+1}$
$=\dfrac{x^2+1}{x^2+x+1}$
b.$E-\dfrac{2}{3}=\dfrac{x^2+1}{x^2+x+1}-\dfrac{2}{3}=\dfrac{x^2-2x+1}{3(x^2+x+1)}=\dfrac{(x-1)^2}{3(x^2+x+1)}$
$\rightarrow E-\dfrac{2}{3}\ge 0\quad\forall x\rightarrow E\ge \dfrac{2}{3}\rightarrow đpcm$
