ĐK: \(x>4,y\ne-2\)
Đặt a=\(\frac{1}{\sqrt{x-4}}\left(a>0\right)\),\(b=\frac{1}{y+2}\)
Vậy \(\left\{{}\begin{matrix}\frac{3}{\sqrt{x-4}}+\frac{4}{y+2}=7\\\frac{5}{\sqrt{x-4}}-\frac{1}{y-2}=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3a+4b=7\\5a-b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3a+4b=7\\20a-4b=16\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}23a=23\\5a-b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)(tm)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-4}}=1\\\frac{1}{y+2}=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x-4}=1\\y+2=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)(tm)
Vậy (x;y)=(5;-1)