Ta có :
\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(\)\(=\left(x^2+x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
\(\) \(=2x^2+2y^2\)
\(\) \(=2\left(x^2+y^2\right)=VP\)
Vậy \(\left(x+y\right)^2+\left(x-y\right)^2=2\left(x^2+y^2\right)\)