\(2x^3-2x^2+2x-x^2+x-1=2x^3-3x^2+2\\ \text{⇔}3x=3\\ \text{⇔}x=1\)
Lần sau ghi rõ đề ra nhé!
\(pt\Leftrightarrow2x^3-2x^2+2x-x^2+x-1=2x^3-3x^2+2\\ \Leftrightarrow3x=3\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1