$x+y=1\Rightarrow x=1-y;y=1-x$
$\Rightarrow A=\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}$
$đặt: \sqrt{x}=a;\sqrt{y}=b\Rightarrow a^2+b^2=1$
$\Rightarrow A=\frac{a^2}{b}+\frac{b^2}{a}=\frac{a^4}{a^2b}+\frac{b^4}{ab^2}\geq \frac{(a^2+b^2)^2}{a^2b+ab^2}$ ( cô - si)
$\Rightarrow A\geq \frac{1}{a^2b+ab^2}$
$có:a^2b+ab^2=ab(a+b)\leq \frac{a^2+b^2}{2\sqrt{2(a^2+b^2)}}$
$\Rightarrow A\geq \frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}$
$\Rightarrow MinA=........; dấu "="$ xảy ra......
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