ĐK: \(x\ge0,y\ge0\)
\(\left\{{}\begin{matrix}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=12\\\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=12\\\left(\sqrt{x}+\sqrt{y}\right)\left[\left(\sqrt{x}+\sqrt{y}\right)^2-3\sqrt{xy}\right]=28\end{matrix}\right.\)(1)
Đặt \(a=\sqrt{x}+\sqrt{y}\left(a\ge0\right)\)
b=\(\sqrt{xy}\left(b\ge0\right)\)
Vậy (1)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}ab=12\\a\left(a^2-3b\right)=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}ab=12\left(3\right)\\a^3-3ab=28\left(4\right)\end{matrix}\right.\)
Lấy (3) thay vào (4) ta được \(a^3-3.12=28\Leftrightarrow a^3=64\Leftrightarrow a=4\Leftrightarrow b=3\)
Vậy ta có \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=4\\\sqrt{xy}=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=9\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=9\end{matrix}\right.\end{matrix}\right.\)
Vậy (x;y)={(9;1);(1;9)}
