\(\left(1\right)\Leftrightarrow\left(x+y\right)^2-1+2xy\left(\frac{1}{x+y}-1\right)=0\)
\(\Leftrightarrow\left(x+y+1\right)\left(x+y-1\right)-2xy.\frac{x+y-1}{x+y}=0\)
\(\Leftrightarrow x+y-1=0\text{ }or\text{ }x+y+1=\frac{2xy}{x+y}\text{ }\left(3\right)\)
\(\left(3\right)\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)=2xy\)
\(\Leftrightarrow x^2+y^2+x+y=0\text{ }\)- vô nghiệm do x+y>0