(x - 2) + 3x2 - 6x = 0
=> (x - 2) + 3x.(x - 2) = 0
=> (x - 2).(1 + 3x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\1+3x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\3x=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{-1}{3}\end{array}\right.\)
Vậy \(x\in\left\{2;\frac{-1}{3}\right\}\)
(x - 2) + 3x2 - 6x = 0
=> ( x - 2) + 3x . (x -2 ) =0
=> (x - 2) . (1 + 3x) =0
+) x - 2 = 0
=> x = 0 + 2 =2
+) 1 + 3x = 0
3x = 0 - 1 = -1
=> x = -1/3