\(N=\frac{x^2-4x+8-x^2-8x-16}{x^2-4x+8}=1-\frac{\left(x+4\right)^2}{\left(x-2\right)^2+4}\le1\)
\(N_{max}=1\) khi \(x=-4\)
\(N=\frac{-x^2+4x-8+x^2}{x^2-4x+8}=-1+\frac{x^2}{\left(x-2\right)^2+4}\ge-1\)
\(N_{min}=-1\) khi \(x=0\)
Lời giải:
\(N=\frac{4(x-2)}{(x^2-4x+4)+4}=\frac{4(x-2)}{(x-2)^2+4}=\frac{4t}{t^2+4}\)
Có:
\(N+2=\frac{t^2+4t+4}{t^2+4}=\frac{(t+2)^2}{t^2+4}\geq 0, \forall t\in\mathbb{R}\)
\(\Rightarrow N\geq -2\) hay $N_{\min}=-2$ khi $t=-2\Leftrightarrow x=0$
\(N-2=-\frac{t^2-4t+4}{t^2+4}=\frac{-(t-2)^2}{t^2+4}\leq 0, \forall t\in\mathbb{R}\)
\(\Rightarrow N\leq 2\) hay $N_{\max}=2$ khi $t=2\Leftrightarrow x=4$
Vậy......
