Phương trình hay là bất phương trình vậy bạn ???
\(\left(x+y\right)^2=\left(x+1\right)\left(y+1\right)\)
\(\Leftrightarrow x^2+2xy+y^2=xy+y+x+1\)
\(\Leftrightarrow x^2-x+xy+y^2-y-1=0\)
\(\Leftrightarrow2x^2-2x+2xy+2y^2-2y-2=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-4=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(x+y-2\right)\left(x+y+2\right)-0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-1=0\\\left[{}\begin{matrix}x+y-2=0\\x+y+2=0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\\x+y-2=0\left(thỏa-mãn\right)\end{matrix}\right.\)
Vậy x=y=1
