[Toán 8] Chứng minh

I

iceghost

Từ $x+y+z=0$
$\implies x^3+y^3+z^3 = 3xyz$
Dễ cm $2(xy+yz+xz) = -(x^2+y^2+z^2)$

Ta có : $(x^2+y^2+z^2)(x^3+y^3+z^3) = x^5+y^5+z^5+x^2y^2(x+y)+y^2z^2(y+z)+x^2z^2(x+z)$
$\iff 3xyz(x^2+y^2+z^2) = x^5+y^5+z^5-x^2y^2z-xy^2z^2-x^2yz^2 \\
\iff 3xyz(x^2+y^2+z^2) = x^5+y^5+z^2-xyz(xy+yz+xz) \\
\iff 6xyz(x^2+y^2+z^2) = 2(x^5+y^5+z^2)-2xyz(xy+yz+xz) \\
\iff 6xyz(x^2+y^2+z^2) = 2(x^5+y^5+z^2)+xyz(x^2+y^2+z^2) \\
\iff 5xyz(x^2+y^2+z^2) = 2(x^5+y^5+z^2)$
 
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