tìm x thuộc z :x^2 - 3x + 2 = 0

x²-3x+2=0

⇒x²-2x-x+2=0

⇒x(x-2)-(x-2)=0

⇒(x-2)(x-1)=0

⇒\(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\) 

Vậy x=2 hoặc x=1