Ta có : \(\left|x-2\right|\ge0\)
\(\Rightarrow\left|x-2\right|+3\ge3\)
\(\Rightarrow\frac{1}{3+\left|x-2\right|}\le\frac{1}{3}\)
\(\Rightarrow10-\frac{1}{3+\left|x-2\right|}\ge\frac{29}{3}\)
Dấu " = " xảy ra khi \(x-2=0\)
\(x=2\)
\(\Rightarrow MIN_D=\frac{29}{3}\) khi \(x=2\)