\(\left(x+y\right)^2+\left(1-x\right)\left(1+y\right)=0\)
\(\Leftrightarrow x^2+2xy+y^2+1+y-x-xy=0\)
\(\Leftrightarrow x^2+y^2+xy-x+y+1=0\)
\(\Leftrightarrow2x^2+2y^2+2xy-2x+2y+2=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta thấy : \(\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
Do vậy, dấu "=" xảy ra : \(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-y\\x=1\\y=-1\end{cases}}\) ( thỏa mãn )
Vậy : \(\left(x,y\right)=\left(1,-1\right)\)
