Question

Tìm x:

\(\left(x-2\right)+3x^2-6x=0\)

Answer 1

(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 \(x\in\left\{2;\frac{-1}{3}\right\}\)

Answer 2

(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