G = { x ∈ Z|(x ² – 3x + 2)(x ² – √3x) = 0 Liệt kê các phần tử sau

`G={x\in ZZ|(x^2-3x+2)(x^2-\sqrt{3}x)=0}`

Ta có: `(x^2-3x+2)(x^2-\sqrt{3}x)=0`

`<=>x(x-\sqrt3)(x-1)(x-2)=0`

`<=>`\(\left[ \begin{array}{l}\ \ \ \ x=0\\x-\sqrt3=0\\\ \ x-1=0\\ \ \ x-2=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}\ x=0\\x=\sqrt3\\\ x=1\\\ x=2\end{array} \right.\) 

Mà `x\in ZZ` nên `G={0;1;2}`