Ta có: \(\left|x-2\right|\ge0\Rightarrow\left|x-2\right|+3\ge3\Rightarrow\frac{1}{\left|x-2\right|+3}\ge\frac{1}{3}\)
Dấu '=' xảy ra khi \(\frac{1}{\left|x-2\right|+3}=\frac{1}{3}\Rightarrow\left|x-2\right|+3=3\Rightarrow x-2=0\Rightarrow x=2\)
Vậy Max\(Max\frac{1}{\left|x-2\right|+3}=\frac{1}{3}\) với \(x=2\)