Question

[Nguyễn Anh Dũng An]

Phân tích đa thức thành nhân tử:\(x^4+y^4+\left(x+y\right)^4\)

Answer 1

[kudo shinichi]

\(x^4+y^4+\left(x+y\right)^4\)

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

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

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

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

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

Sai thì thôi nhé~

Answer 2

[Đoàn Đức Hà]

       \(x^4+y^4+\left(x+y\right)^4\)

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

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

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

\(=2\left[\left(x^4+2x^3y+x^2y^2\right)+2\left(x^2+xy\right)y^2+y^4\right]\)

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

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