a/d bunhiacopxki co:
\(S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\le\left(1^2+1^2\right)\left(x-2+y-3\right)=2\cdot1=2\)
\(\Rightarrow S\le\sqrt{2}\)
Dấu ''='' xảy ra khi \(x=\dfrac{5}{2};y=\dfrac{7}{2}\)
Vậy GTLN của S = \(\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\)