Question

Chứng minh đẳng thức:

(x^2+y^2)^2-4x^2y^2=(x+y)^2(x-y)^2

Làm hộ mình với. Thanks

Answer 1

Ta có:

\(\left(x^2+y^2\right)^2-4x^2y^2-\left(x+y\right)^2\left(x-y\right)^2.\)

\(=x^4+2.x^2.y^2+y^4-4x^2y^2-\left[\left(x+y\right)\left(x-y\right)\right]^2\)

\(=x^4+2.x^2.y^2+y^4-4x^2y^2-\left[x^2-y^2\right]^2\)

\(=x^4+2x^2y^2+y^4-4x^2y^2-\left(x^4-2x^2y^2+y^4\right)\)

\(=x^4+2x^2y^2+y^4-4x^2y^2-x^4+2x^2y^2-y^4\)

\(=0\)

Vậy \(\left(x^2+y^2\right)^2-4x^2y^2=\left(x+y\right)^2\left(x-y\right)^2.\)