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\(A=\sqrt{x-1}+\sqrt{y-2}\)
\(\Rightarrow A^2\le2\left(x-1+y-2\right)=2\)
\(\Rightarrow A_{max}=\sqrt{2}\) khi \(\left\{{}\begin{matrix}x-1=y-2\\x+y=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\y=\frac{5}{2}\end{matrix}\right.\)
Áp dụng bất đẳng thức phụ \(a+b\le\sqrt{2\left(a^2+b^2\right)}\left(a+b\ge0\right)\) ta có :
\(S=\sqrt{x-1}+\sqrt{y-2}\le\sqrt{2\left(x-1+y-2\right)}=\sqrt{2}\)
max \(S=\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}x-1=y-2\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\y=\frac{5}{2}\end{matrix}\right.\)
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