Question

Cho x+y+z=0 và x^2+y^2+z^2=9. Tính P=x^4+y^4+z^4

Answer 1

\(x+y+z=0\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)\(=0\)

\(\Rightarrow2xy+2yz+2xz=-9\)

\(\Rightarrow xy+yz+xz=-\frac{9}{2}\)

\(\Rightarrow x^2y^2+y^2z^2+x^2z^2+2xy^2z+2xyz^2+2x^2yz=\left(-\frac{9}{2}\right)^2=\frac{81}{4}\)\(\)

\(\Rightarrow x^2y^2+y^2z^2+x^2z^2+2xyz\left(x+y+z\right)=\frac{81}{4}\)

\(\Rightarrow x^2y^2+y^2z^2+x^2z^2=\frac{81}{4}\)

\(\left(x^2+y^2+z^2\right)^2=9^2=81\)

\(\Rightarrow P=x^4+y^4+z^4=81-2\left(x^2y^2+y^2z^2+x^2z^2\right)=81-2.\frac{81}{4}=\frac{81}{2}\)