\(\frac{x^2-4x-4}{x^2-4x+5}=\frac{x^2-4x+5-9}{x^2-4x+5}=1-\frac{9}{x^2-4x+5}=1-\frac{9}{\left(x-2\right)^2+1}\ge1-9=-8\)\("="\Leftrightarrow x=2\)
\(=\frac{x^2-4x+5-9}{x^2-4x+5}=1-\frac{9}{x^2-4x+4+1}=1-\frac{9}{\left(x-2\right)^2+1}\ge1-\frac{9}{1}=-8\)
vậy MIN =-8 với x=2