Do \(a\ge2\Rightarrow\sqrt{a-2}\ge0\)
\(b\ge3\Rightarrow\sqrt{b-3}\ge0\)
\(\Rightarrow\sqrt{a-2}+\sqrt{b-3}\ge0\)
Dấu ''='' xảy ra \(\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)
Vậy GTNN \(A=0\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)