a, \(E=1+\left(\frac{2x^3+x^2-x}{x^3-1}-\frac{2x-1}{x-1}\right)\frac{x^2-x}{2x-1}\)
\(=1+\left[\frac{x\left(x+1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x-1}{x-1}\right]\frac{x\left(x-1\right)}{2x-1}\)
\(=1+\frac{x\left(x+1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x\left(x+1\right)}{2x-1}-\frac{2x-1}{x-1}.\frac{x\left(x-1\right)}{2x-1}\)
\(=1+x.\frac{x\left(x+1\right)}{x^2+x+1}-x\)
\(=1+x\left(\frac{x^2+x}{x^2+x+1}-1\right)\)
\(=1+x.\frac{x^2+x-x^2-x-1}{x^2+x+1}\)
\(=1+x.\frac{-1}{x^2+x+1}\)
\(=1-\frac{x}{x^2+x+1}\)
\(=\frac{x^2+1}{x^2+x+1}\)
b, \(E-\frac{2}{3}=\frac{x^2+1}{x^2+x+1}-\frac{2}{3}=\frac{x^2-2x+1}{3\left(x^2+x+1\right)}=\frac{\left(x-1\right)^2}{3\left(x^2+x+1\right)}\)
\(\Rightarrow E-\frac{2}{3}\ge0\forall x\Rightarrow E\ge\frac{2}{3}\left(đpcm\right)\)
