Tìm x,biết: (a) 3x(x−2)=8(x−2). (b) 2x(x−4)−x^2+16=0.

a) \(\begin{array}{l}3x\left( {x - 2} \right) = 8\left( {x - 2} \right)\\3x\left( {x - 2} \right) = 8\left( {x - 2} \right)\\\left( {x - 2} \right)\left( {3x - 8} \right) = 0\end{array}\)

\(x - 2 = 0\) hoặc \(3x - 8 = 0\)

\(x = 2\) hoặc \(x = \frac{8}{3}\)

Vậy \(x \in \left\{ {2;\frac{8}{3}} \right\}\).

b) \(\begin{array}{l}2x(x - 4) - {x^2} + 16 = 0\\2x(x - 4) - \left( {{x^2} - 16} \right) = 0\\2x(x - 4) - (x - 4)(x + 4) = 0\\(x - 4)\left( {2x - x - 4} \right) = 0\\{(x - 4)^2} = 0\\x - 4 = 0\\x = 4\end{array}\)

Vậy \[x = 4\].