Theo bài ra ta có:
x + y +z = 0 => \(\left(x+y+z^{ }\right)^2\)=0
<=> \(x^2+y^2+z^2+2.\left(xy+yz+xz\right)=0\)
<=>1+ 2.\(\left(xy+yz+xz\right)\)=0
<=> 2.\(\left(xy+yz+xz\right)\)=-1
<=> \(xy+yz+xz=\frac{-1}{2}\)
=> \(\left(xy+yz+xz\right)\)^2= \(\frac{1}{4}\)
<=> \(\left(xy\right)^2\) \(+\left(yz\right)^2+\left(xz\right)^2+2.xyz.\left(x+y+z\right)=\frac{1}{4}\) (1)
Thay \(x+y+z=0\)ta được: \(2.xyz.\left(x+y+z\right)=0\)
=> Pt (1) viết thành:
\(\left(xy\right)^2\)\(+\left(yz\right)^2+\left(xz\right)^2=\frac{1}{4}\) (*)
Ta có: \(x^2+y^2+z^2=1\)=> \(\left(x^2+y^2+z^2\right)^2=1\)
<=> \(x^4+y^4+z^4\) \(+2.\left[\left(xy\right)^2+\left(yz\right)^2+\left(xz\right)^2\right]\) =1 (2)
Thay (*) vào pt (2) ta được:
\(x^4+y^4+z^4\)+ 2.\(\frac{1}{4}=1\)=> \(x^4+y^4+z^4\)=\(\frac{1}{2}\) => 2.(\(x^4+y^4+z^4\))=1
Vậy M=1
