ĐK: \(x+y>0\)
Đặt \(\sqrt{x+y}=a,x-y=b\left(a>0\right)\)
Hệ\(\Leftrightarrow\hept{\begin{cases}b^2+\frac{a^4-b^2}{a^2}\\a+b=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(a^2-1\right)\left(a^2+b^2\right)=0\\a+b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2=1\\a+b=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=1\\b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y=1\\x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\end{cases}}}\)
Vậy.........
