Ta có:-x²+8x-12=0
⇔ -(x²-8x+12)=0
⇔ x²-8x+12=0
⇔ x²-2x-6x+12=0
⇔ (x²-2x)-(6x-12)=0
⇔ x.(x-2)-6.(x-2)=0
⇔ (x-6).(x-2)=0
⇔\(\left[ \begin{array}{l}x-6=0\\x-2=0\end{array} \right.\)⇔ \(\left[ \begin{array}{l}x=6\\x=2\end{array} \right.\)
Vậy X ∈{2;6}
-----------------------Nguyễn Hoạt--------------------------