x² + y² = 36 - 2xy 

=> ( x +y )² = 36 

=> x+y = 6 or -6

x²-y²= 12 

=> (x+y)(x-y) =12 

If x+y = 6 (1)

=> x-y = 2 (2)

(1),(2)=> x=4;y=2 => xy=6

If x+y = -6 (1)

=> x-y = -2 (2)

(1),(2)=> x=-4; y=-2

=> xy = 6

x2 + y2 = 36 - 2xy 

=> ( x +y )2 = 36 

=> x+y = 6 or -6

x2-y2= 12 

=> (x+y)(x-y) =12 

If x+y = 6 (1)

=> x-y = 2 (2)

(1),(2)=> x=4;y=2 => xy=6

If x+y = -6 (1)

=> x-y = -2 (2)

(1),(2)=> x=-4; y=-2

=> xy = 6

Ngô Phương: x = 4, y= 2 => xy = 6 ?

x² + y² = 36 - 2xy

=> x² + 2xy + y² = 36

=> (x+y)² = 36

=> x + y = 6 or -6

x² - y² = 12 

=> (x+y)(x-y) = 12

=> \(\left[{}\begin{matrix}6\left(x-y\right)=12\\-6\left(x-y\right)=12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-y=2\\x-y=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=6\\x-y=2\end{matrix}\right.\Rightarrow x=4;y=2}\\\left\{{}\begin{matrix}x+y=-6\\x-y=-2\end{matrix}\right.\Rightarrow x=-4;y=-2\end{matrix}\right.\)

So xy = 8