Đáp án + Giải thích các bước giải:
$(3x-1)(x-2)-(2x-1)^2=x(x-2)$
$⇔ 3x^2-6x-x+2-(4x^2-4x+1)=x^2-2x$
$⇔ 3x^2-6x-x+2-4x^2+4x-1=x^2-2x$
$⇔ 3x^2-6x-x+2-4x^2+4x-1-x^2+2x=0$
$⇔ (3x^2-4x^2-x^2)+(-6x-x+4x+2x)+(2-1)=0$
$⇔ -2x^2-x+1=0$
$⇔ -2x^2+x-2x+1=0$
$⇔ -x(2x-1)-(2x-1)=0$
$⇔ (-x-1)(2x-1)=0$
\(⇔ \left[ \begin{array}{l}-x-1=0\\2x-1=0\end{array} \right.\)
\(⇔ \left[ \begin{array}{l}x=-1\\x=\dfrac12\end{array} \right.\)
Vậy $S=\left\{-1;\dfrac12\right\}$
