Giải thích các bước giải:
$\text{Bài 1:}$
$a)2x^2-4x+2$
$=2(x^2-2x+1)$
$=2(x-1)^2$
$b)x^2-y^2+3x-3y$
$=(x+y)(x-y)+3(x-y)$
$=(x-y)(x+y+3)$
$\text{Bài 2:}$
$b)3x(x-1)=1-x$
$⇔3x(x-1)+x-1=0$
$⇔(3x+1)(x-1)=0$
$⇔\left[ \begin{array}{l}3x+1=0\\x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=-\dfrac{1}{3}\\x=1\end{array} \right.$
$\text{Vậy $x∈\{-\dfrac{1}{3};1\}$}$
Học tốt!!!