Question

Cho x;y;z thỏa mãn \(x+y+z=0\) và \(x^2+y^2+z^2=a^2\)

Tính \(x^4+y^4+z^4\)theo a

Answer 1

\(x+y+z=0< =>x+y=-z=>\left(x+y\right)^2=\left(-z\right)^2.\)

\(< =>x^2+2xy+y^2=z^2< =>x^2+y^2-z^2=-2xy\)

\(< =>\left(x^2+y^2-z^2\right)=\left(-2xy\right)^2\)

\(< =>x^4+y^4+z^4+2x^2y^2-2x^2z^2-2y^2z^2=4x^2y^2\)

\(< =>x^4+y^4+z^4=2x^2y^2+2y^2z^2+2x^2z^2\)

\(< =>2\left(x^4+y^4+z^4\right)=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2=\left(x^2+y^2+z^2\right)^2.\)

\(< =>x^4+y^4+z^4=\frac{\left(x^2+y^2+z^2\right)^2}{2}=\frac{a^4}{2}\)

Vậy \(x^4+y^4+z^4=\frac{a^4}{2}\)