Giải:\(\eqalign{
& f'\left( x \right) = 1 - {{2x} \over {\sqrt {{x^2} + 12} }} \le 0{\rm{ }} \cr
& \Leftrightarrow \sqrt {{x^2} + 12} \le 2x \cr
& \Leftrightarrow \left\{ \matrix{
{x^2} + 12 \le 4{x^2} \hfill \cr
x \ge 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
3{x^2} \ge 12 \hfill \cr
x \ge 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
{x^2} \ge 4 \hfill \cr
x \ge 0 \hfill \cr} \right. \Leftrightarrow x \ge 2. \cr}\) Đáp số: \({\rm{[}}2; + \infty ).\)
