Đặt \(\left\{{}\begin{matrix}x\sqrt{y}=a\\y\sqrt{x}=b\end{matrix}\right.\)
Hpt \(\Leftrightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^2=20\end{matrix}\right.\)
=> Hệ đối xứng loại 1 => EZ
Đặt \(\left\{{}\begin{matrix}a=x\sqrt{y}\\b=\sqrt{x}.y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^2=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^2+\left(6-a\right)^2=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\2a^2-12a+16=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\\left[{}\begin{matrix}a=4\\b=2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\\\left\{{}\begin{matrix}a=2\\b=4\end{matrix}\right.\end{matrix}\right.\)
Trường hợp \(\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{y}=4\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{y}=2\sqrt{x}.y\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{y}-2\sqrt{x}.y=0\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}-2\sqrt{y}\right)=0\\y\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\sqrt{y}\\\sqrt{x}.y=2\end{matrix}\right.\)( vì \(\sqrt{xy}\ne0\) )
\(\Leftrightarrow\left\{{}\begin{matrix}x=4y\\\sqrt{4y}.y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4y\\y\sqrt{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4\end{matrix}\right.\)
TRường hợp 2 tương tự nha