(x^2 – 4) + (x – 2)(3 – 2x) = 0

`(x^2 -4) +(x-2)(3-2x)=0`

`⇔(x-2)(x+2) +(x-2)(3-2x)=0`

`⇔(x-2).(x+2+3-2x)=0`

`⇔(x-2)(5-x)=0`

$⇔\left[\begin{matrix} x-2=0\\ 5-x=0\end{matrix}\right.$

$⇔\left[\begin{matrix} x=2\\ x=5\end{matrix}\right.$

Vậy `S={2;5}`

#andy

$\begin{array}{l} \left( {{x^2} - 4} \right) + \left( {x - 2} \right)\left( {3 - 2x} \right) = 0\\  \Leftrightarrow \left( {x - 2} \right)\left( {x + 2} \right) + \left( {x - 2} \right)\left( {3 - 2x} \right) = 0\\  \Leftrightarrow \left( {x - 2} \right)\left( {x + 2 + 3 - 2x} \right) = 0\\  \Leftrightarrow \left( {x - 2} \right)\left( { - x + 5} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\  - x + 5 = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = 2\\  - x =  - 5 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 5 \end{array} \right.\\  \Rightarrow S = \left\{ {2;5} \right\} \end{array}$