Question

[Trần Thị Kiều Trâm]

\(CMR\)nếu \(x+y+z=0\)thì:

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

Answer 1

[ST]

Ta có: x+y+z=0 => x³+y³+z³=3xyz (tự c/m)

Mặt khác \(x+y+z=0\Leftrightarrow x+y=-z\Leftrightarrow\left(x+y\right)^2=\left(-z\right)^2\)

\(\Leftrightarrow x^2+2xy+y^2=z^2\Leftrightarrow x^2+y^2=z^2-2xy\)

Tương tự ta cũng có: \(y^2+z^2=x^2-2yz;z^2+x^2=y^2-2zx\)

Lại có: \(\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)

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

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

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

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

\(\Rightarrow3xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^5\right)-2xyz\left(x^2+y^2+z^2\right)\)

\(\Rightarrow5xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^5\right)\) (đpcm)

Answer 2

[Huy]

I love 💑