Ta có: \(\left(x^2+1\right)\left(y^2+1\right)+2\left(x-y\right)\left(1-xy\right)=4\left(1+xy\right)\)
\(\Leftrightarrow x^2y^2+x^2+y^2+1-2\left(x-y\right)\left(xy-1\right)=4+4xy\)
\(\Leftrightarrow\left(x^2y^2-2xy+1\right)+\left(x^2-2xy+y^2\right)-2\left(x-y\right)\left(xy-1\right)=4\)
\(\Leftrightarrow\left(xy-1\right)^2-2\left(x-y\right)\left(xy-1\right)+\left(x-y\right)^2=4\)
\(\Leftrightarrow\left(xy-1-x+y\right)^2=4\)
\(\Leftrightarrow\left[\left(x+1\right)\left(y-1\right)\right]^2=4\)
\(\Leftrightarrow\left(x+1\right)^2\left(y-1\right)^2=4=1.4\)
Vì \(\left(x+1\right)^2;\left(y-1\right)^2\) là các SCP và đều không âm nên ta chỉ cần xét các TH sau:
TH1: \(\hept{\begin{cases}\left(x+1\right)^2=1\\\left(y-1\right)^2=4\end{cases}}\) => \(\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\) và \(\orbr{\begin{cases}y-1=2\\y-1=-2\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\) và \(\orbr{\begin{cases}y=3\\y=-1\end{cases}}\)
TH2: \(\hept{\begin{cases}\left(x+1\right)^2=4\\\left(y-1\right)^2=1\end{cases}}\) => \(\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\) và \(\orbr{\begin{cases}y-1=1\\y-1=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\) và \(\orbr{\begin{cases}y=2\\y=0\end{cases}}\)
Kết luận:...
\(\left(x^2+1\right)\left(y^2+1\right)+2\left(x-y\right)\left(1-xy\right)=4\left(1+xy\right)\)
\(\Leftrightarrow\left(1-2xy+x^2y^2\right)+2\left(x-y\right)\left(1-xy\right)=4+4xy\)
\(\Leftrightarrow\left(1-xy\right)^2+2\left(x-y\right)\left(1-xy\right)+\left(x^2-2xy+y^2\right)=4\)
\(\Leftrightarrow\left(1-xy\right)^2+2\left(x-y\right)\left(1-xy\right)+\left(x-y\right)^2=4\)
\(\Leftrightarrow\left(1-xy+x-y\right)^2=4\)
\(\Leftrightarrow\left[\left(x+1\right)\left(1-y\right)\right]^2=2^2\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)\left(1-y\right)=2\\\left(x+1\right)\left(1-y\right)=-2\end{cases}}\)
Tự xét các TH
