ĐKXĐ: \(x,y\ge0\).
Đặt \(\sqrt{x}=a;\sqrt{y}=b\left(a,b\ge0\right)\).
HPT đã cho tương đương với:
\(\left\{{}\begin{matrix}a+b+4ab=16\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+\left(2a^2+2b^2+4ab\right)=36\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(a+b\right)^2+\left(a+b\right)-36=0\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b-4\right)\left[2\left(a+b\right)+9\right]=0\\a^2+b^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\) (Do a + b \(\ge\) 0).
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\ab=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=1\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1;y=9\\x=9;y=1\end{matrix}\right.\).
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