Lời giải:
Đặt \(\sqrt{3-x}=a (a\geq 0)\Rightarrow x=3-a^2\)
Khi đó:
\(y=a+3-a^2=\frac{13}{4}-(a^2-a+\frac{1}{4})\)
\(=\frac{13}{4}-(a-\frac{1}{2})^2\leq \frac{13}{4}\)
Vậy \(y_{\max}=\frac{13}{4}\Leftrightarrow a=\frac{1}{2}\Leftrightarrow x=\frac{11}{4}\)