\(\hept{\begin{cases}x^2+8=xy^2+2x\left(1\right)\\y^2+8=x^2y+2y\left(2\right)\end{cases}}\)
\(\left(1\right)-\left(2\right)\Leftrightarrow\left(x-y\right)\left(x+y\right)=-xy\left(x-y\right)+2\left(x-y\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+xy-2\right)=0\)
Đến đây dễ r :)))