\(A\le\sqrt{2\left(x-1+y-2\right)}=\sqrt{2\left(x+y-3\right)}=\sqrt{2}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{3}{2}\\y=\frac{5}{2}\end{matrix}\right.\)
\(A^2=\left(\sqrt{x-1}.1+\sqrt{y-2}.1\right)^2\le\left(1+1\right)\left(x-1+y-2\right)=2.1=2\Rightarrow A\le\sqrt{2}\).
Đẳng thức xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x+y=4\\\frac{\sqrt{x-1}}{1}=\frac{\sqrt{y-2}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{6}{5}\\y=\frac{14}{5}\end{matrix}\right.\).
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