\(A=\dfrac{x^2-2x+2x-4+4}{x-2}=\dfrac{x\left(x-2\right)+2\left(x-2\right)+4}{x-2}=x+2+\dfrac{4}{x-2}\)
\(=x-2+\dfrac{4}{x-2}+4\ge2\sqrt{\dfrac{\left(x-2\right).4}{x-2}}+4=8\)
Dấu ''='' xảy ra khi \(\left(x-2\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)