Question

rut gon bieu thuc

p=(x^2+2xy)^2+2(x^2+2xy)y^2+y^4

Answer 1

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

\(=x^4+4x^3y+4x^2y^2+2x^2y^2+4xy^3+y^4\)

\(=x^4+y^4+6x^2y^2+4x^3y+4xy^3\)

Answer 2

P = ( x² + 2xy )² + 2( x² + 2xy )y² + y⁴

= ( x² + 2xy )² + 2( x² + 2xy )y² + ( y² )²

= ( x² + 2xy + y² )²

= [ ( x + y )² ]²

= ( x + y )⁴

Answer 3

            Bài làm :

Ta có :

P = ( x² + 2xy )² + 2( x² + 2xy )y² + y⁴

P = ( x² + 2xy )² + 2( x² + 2xy )y² + ( y² )²

P = ( x² + 2xy + y² )²

P = [ ( x + y )² ]²

P = ( x + y )⁴

Vậy P = (x + y)⁴

Answer 4

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

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