Question

[Ngô Đức Duy]

Cho x+y+z=0. CM:

\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)

Answer 1

[ST]

Ta có: \(x+y+z=0\Leftrightarrow x+y=-z\)

\(\Leftrightarrow\left(x+y\right)^3=-z^3\Leftrightarrow x^3+3xy\left(x+y\right)+y^3=-z^3\)

\(\Leftrightarrow x^3-3xyz+y^3=-z^3\Leftrightarrow x^3+y^3+z^3=3xyz\)

Do đó \(3xyz\left(x^2+y^2+z^2\right)=\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)

\(=x^5+y^5+z^5+x^3\left(y^2+z^2\right)+y^3\left(z^2+x^2\right)+z^3\left(x^2+y^2\right)\) (*)

Mà \(x^2+y^2=\left(x+y\right)^2-2xy=\left(-z\right)^2-2xy=z^2-2xy\) (vì x + y = -z) (1)

Tương tự, ta có: \(y^2+z^2=x^2-2yz\left(2\right);z^2+x^2=y^2-2zx\left(3\right)\)

Thay (1);(2);(3) vào (*) ta được:

\(3xyz\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3\left(x^2-2yz\right)+y^3\left(y^2-2zx\right)+z^3\left(z^2-2xy\right)\)

\(=x^5+y^5+z^5+x^5-2x^3yz+y^5-2xy^3z+z^5-2xyz^3\)

\(=2\left(x^5+y^5+z^5\right)-2xyz\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow3xyz\left(x^2+y^2+z^2\right)+2xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^2\right)-2xyz\left(x^2+y^2+z^2\right)+2xyz\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow5xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^5\right)\left(đpcm\right)\)

Answer 2

[Nguyễn Tũn]

tích mình đi

ai tích mình

mình ko tích lại đâu

thanks