Question

[trần xuân quyến]

cho x,y,z >0 thỏa mãn \(x^3+y^3+z^3=3xyz\)

tính \(P=\frac{\left(x-y\right)^{2017}}{x+y}+\frac{\left(y-z\right)^{2018}}{y+z}+\frac{\left(z-x\right)^{2019}}{x+z}\)

Answer 1

[Nguyễn Anh Quân]

x^3+y^3+z^3-3xyz = 0

<=> (x+y+z).(x^2+y^2+z^2-xy-yz-zx) = 0

Mà x+y+z > 0 => x^2+y^2+z^2-xy-yz-zx = 0

<=> 2x^2+2y^2+2z^2-2xy-2yz-2zx = 0

<=> (x-y)^2+(y-z)^2+(z-x)^2 = 0

=> x-y=0;y-z=0;z-x=0

=> P = 0

k mk nha