ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{3x+y}=a\ge0\\\sqrt{5x+4y}=b\ge0\end{matrix}\right.\) \(\Rightarrow2a^2-b^2=x-2y\)
Hệ trở thành: \(\left\{{}\begin{matrix}a+b=5\\12b+2a^2-b^2=35\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=5-b\\12b+2a^2-b^2=35\end{matrix}\right.\)
\(\Rightarrow12b+2\left(5-b\right)^2-b^2=25\)
\(\Rightarrow...\)