Câu 4 :Ta có : x + y + z = 0 => ( x+y+z )^2 = 0<=> x^2 + y^2 + z^2 + 2( xy+yz+xz ) = 0 <=> xy + yz + xz = -a^2/2Để ý rằng :x+4 + y^4 + z^4 = ( x^2 + y^2 + z^2 )^2 - 2( x^2y^2 + y^2z^2 + z^2x^2 ) = a^4 - 2[( xy + yz + xz)^2 - 2xyz( x+y+z )]= a^4 -2(xy + yz + xz)^2 = a^4 - 2 . a^4/4 = a^4/2
Câu 4 :Ta có : x + y + z = 0 =>
$( x+y+z )^2
$ = 0<=>
$x^2 + y^2 + z^2
$ + 2( xy+yz+xz ) = 0 <=> xy + yz + xz =
$ \frac{-
2^
{2
}}{2
}$ Để ý rằng :x+4 +
$y^4 + z^4 = ( x^2 + y^2 + z^2 )^2 - 2( x^2y^2 + y^2z^2 + z^2x^2 )
$ =
$a^4 - 2[( xy + yz + xz)^2 - 2xyz( x+y+z )]
$=
$a^4 -2(xy + yz + xz)^2
$ =
$a^4
$ - 2 .
$ \frac{a^
{4
}}{4
} =
\frac{a^
{4
}}{2
}$