Đặt \(A=\frac{\sqrt{x}}{\sqrt{x}+3}\) . Nhận xét : A > 0
\(A=\frac{\sqrt{x}}{\sqrt{x}+3}=\frac{\sqrt{x}+3-3}{\sqrt{x}+3}=1-\frac{3}{\sqrt{x}+3}\)
Ta luôn có : \(\sqrt{x}+3\ge3\Leftrightarrow\frac{3}{\sqrt{x}+3}\le1\Leftrightarrow-\frac{3}{\sqrt{x}+3}\ge-1\)
Hay \(A\ge0\)
Vậy MinA = 0 <=> x = 0