Giải:
\(\left(x+1\right)\left(x^2-x+1\right)-2x=x\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^3+1-2x=x\left(x^2-4\right)\)
\(\Leftrightarrow x^3+1-2x=x^3-4x\)
\(\Leftrightarrow x^3+1-2x-x^3+4x=0\)
\(\Leftrightarrow1+2x=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(\left(x+1\right)\left(x^2-x+1\right)-2x=x\left(x+2\right)\left(x-2\right)\)
\(\Leftrightarrow x^3-1-2x=x\left(x^2-4\right)\)
\(\Leftrightarrow x^3-1-2x=x^3-4x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)